{"id":21108,"date":"2026-01-10T14:34:31","date_gmt":"2026-01-10T06:34:31","guid":{"rendered":"https:\/\/3d-universal.com\/en\/?p=21108"},"modified":"2026-01-10T14:37:50","modified_gmt":"2026-01-10T06:37:50","slug":"momentum-and-collisions","status":"publish","type":"post","link":"https:\/\/3d-universal.com\/en\/blogs\/momentum-and-collisions.html","title":{"rendered":"Momentum and Collisions: NMAT Physics Review"},"content":{"rendered":"<p><!--more--><\/p>\n<h1>Momentum and Collisions: NMAT Physics Review<\/h1>\n<p>Momentum and collisions are core topics in NMAT Physics that frequently appear in conceptual questions,<br \/>\nnumerical problems, and real-life applications. A strong understanding of momentum, impulse, and collision<br \/>\ntypes allows NMAT examinees to solve problems efficiently and avoid common traps. This review provides a<br \/>\ncomprehensive explanation of momentum principles, conservation laws, collision analysis, and NMAT-focused<br \/>\nproblem-solving strategies.<\/p>\n<h2>What Is Momentum?<\/h2>\n<p>Momentum is a vector quantity that describes the motion of an object. It depends on both mass and velocity,<br \/>\nmaking it a more complete measure of motion than velocity alone.<\/p>\n<p><strong>Definition:<\/strong><\/p>\n<p>Momentum (<em>p<\/em>) is defined as:<\/p>\n<p><strong>p = mv<\/strong><\/p>\n<p>where <em>m<\/em> is the mass of the object and <em>v<\/em> is its velocity.<\/p>\n<p>Because velocity is a vector, momentum also has both magnitude and direction. This directional nature is<br \/>\ncritical when solving collision problems involving multiple objects.<\/p>\n<h2>Units of Momentum<\/h2>\n<p>In the International System of Units (SI), momentum is measured in:<\/p>\n<p><strong>kg\u00b7m\/s<\/strong><\/p>\n<p>Momentum can also be expressed as newton-seconds (N\u00b7s), since:<\/p>\n<p>1 N\u00b7s = 1 kg\u00b7m\/s<\/p>\n<p>This equivalence becomes important when discussing impulse, which is closely related to momentum.<\/p>\n<h2>Impulse and Its Relationship to Momentum<\/h2>\n<p>Impulse describes the effect of a force acting over a time interval. In many NMAT problems, impulse is used<br \/>\nto analyze collisions and sudden changes in motion.<\/p>\n<p><strong>Impulse (J):<\/strong><\/p>\n<p>J = F\u0394t<\/p>\n<p>where <em>F<\/em> is the applied force and <em>\u0394t<\/em> is the time interval during which the force acts.<\/p>\n<p>Impulse is directly related to the change in momentum:<\/p>\n<p><strong>J = \u0394p = mv<sub>f<\/sub> \u2212 mv<sub>i<\/sub><\/strong><\/p>\n<p>This relationship is known as the impulse-momentum theorem. It is frequently tested in NMAT physics questions<br \/>\nthat involve stopping distances, safety devices, or impact forces.<\/p>\n<h2>Law of Conservation of Momentum<\/h2>\n<p>The law of conservation of momentum is one of the most important principles in collision problems.<\/p>\n<p><strong>Statement:<\/strong><\/p>\n<p>If no external force acts on a system, the total momentum of the system remains constant.<\/p>\n<p>Mathematically:<\/p>\n<p>Total momentum before interaction = Total momentum after interaction<\/p>\n<p>This law applies to all types of collisions\u2014elastic, inelastic, and perfectly inelastic\u2014as long as the system<br \/>\nis isolated.<\/p>\n<h2>Conditions for Momentum Conservation<\/h2>\n<p>Momentum is conserved when:<\/p>\n<ul>\n<li>The system is isolated (no external forces)<\/li>\n<li>External forces are negligible compared to internal forces<\/li>\n<li>The time interval of interaction is very short<\/li>\n<\/ul>\n<p>In NMAT problems, surfaces are often assumed to be frictionless, and air resistance is neglected to allow<br \/>\nmomentum conservation.<\/p>\n<h2>Types of Collisions<\/h2>\n<p>Collisions are interactions between objects that occur over a very short time interval. They are classified<br \/>\nbased on whether kinetic energy is conserved.<\/p>\n<h2>Elastic Collisions<\/h2>\n<p>In an elastic collision:<\/p>\n<ul>\n<li>Momentum is conserved<\/li>\n<li>Kinetic energy is conserved<\/li>\n<\/ul>\n<p>Elastic collisions are idealized and rarely occur in everyday life, but they are common in physics problems,<br \/>\nespecially those involving particles or ideal objects.<\/p>\n<p>Example systems include gas molecules and idealized billiard balls.<\/p>\n<h2>Inelastic Collisions<\/h2>\n<p>In an inelastic collision:<\/p>\n<ul>\n<li>Momentum is conserved<\/li>\n<li>Kinetic energy is not conserved<\/li>\n<\/ul>\n<p>Some kinetic energy is transformed into heat, sound, or deformation. Most real-world collisions fall into<br \/>\nthis category.<\/p>\n<h2>Perfectly Inelastic Collisions<\/h2>\n<p>A perfectly inelastic collision is a special case of inelastic collision where:<\/p>\n<ul>\n<li>The objects stick together after the collision<\/li>\n<li>Momentum is conserved<\/li>\n<li>Kinetic energy loss is maximum<\/li>\n<\/ul>\n<p>These problems are very common in NMAT exams because they involve straightforward momentum conservation and<br \/>\nsimple algebra.<\/p>\n<h2>One-Dimensional Collisions<\/h2>\n<p>In one-dimensional (1D) collisions, all motion occurs along a straight line. These are the most common type<br \/>\nof collision problems in NMAT.<\/p>\n<p>For two objects:<\/p>\n<p>m<sub>1<\/sub>u<sub>1<\/sub> + m<sub>2<\/sub>u<sub>2<\/sub> = m<sub>1<\/sub>v<sub>1<\/sub> + m<sub>2<\/sub>v<sub>2<\/sub><\/p>\n<p>where:<\/p>\n<ul>\n<li>u = initial velocity<\/li>\n<li>v = final velocity<\/li>\n<\/ul>\n<p>Always choose a consistent direction as positive to avoid sign errors.<\/p>\n<h2>Two-Dimensional Collisions<\/h2>\n<p>In two-dimensional (2D) collisions, objects move in different directions after impact. Momentum must be<br \/>\nconserved separately in each perpendicular direction.<\/p>\n<p>Momentum conservation equations:<\/p>\n<ul>\n<li>Total momentum in x-direction before = after<\/li>\n<li>Total momentum in y-direction before = after<\/li>\n<\/ul>\n<p>These problems often involve vector components and trigonometry, which can be challenging but are less<br \/>\nfrequently tested in NMAT compared to 1D collisions.<\/p>\n<h2>Center of Mass and Momentum<\/h2>\n<p>The center of mass of a system behaves as if all the system\u2019s mass were concentrated at that point.<\/p>\n<p>The velocity of the center of mass remains constant if no external force acts on the system.<\/p>\n<p>This concept reinforces the conservation of momentum and helps explain collision behavior in multi-object<br \/>\nsystems.<\/p>\n<h2>Recoil and Explosion Problems<\/h2>\n<p>Recoil and explosion problems are applications of momentum conservation.<\/p>\n<p>In an explosion:<\/p>\n<ul>\n<li>The initial momentum is usually zero<\/li>\n<li>Fragments move apart with equal and opposite total momentum<\/li>\n<\/ul>\n<p>These problems are common in NMAT and often test conceptual understanding rather than heavy computation.<\/p>\n<h2>Applications of Momentum and Collisions<\/h2>\n<p>Momentum concepts are widely applied in real life:<\/p>\n<ul>\n<li>Vehicle safety systems (airbags, seatbelts)<\/li>\n<li>Sports collisions<\/li>\n<li>Rocket propulsion<\/li>\n<li>Ballistics and impact analysis<\/li>\n<\/ul>\n<p>NMAT questions often describe real-world scenarios to test conceptual application rather than formula<br \/>\nmemorization.<\/p>\n<h2>Common NMAT Traps and Mistakes<\/h2>\n<p>NMAT physics questions are designed to test clarity of understanding. Common mistakes include:<\/p>\n<ul>\n<li>Forgetting that momentum is a vector<\/li>\n<li>Assuming kinetic energy is conserved in all collisions<\/li>\n<li>Ignoring direction signs<\/li>\n<li>Applying momentum conservation when external forces are present<\/li>\n<\/ul>\n<p>Carefully reading the problem statement and identifying the type of collision is essential.<\/p>\n<h2>NMAT Problem-Solving Strategy<\/h2>\n<p>To solve momentum and collision problems efficiently:<\/p>\n<ol>\n<li>Identify the system and check if momentum is conserved<\/li>\n<li>Determine the type of collision<\/li>\n<li>Choose a reference direction<\/li>\n<li>Write momentum equations clearly<\/li>\n<li>Solve algebraically before substituting numbers<\/li>\n<\/ol>\n<p>Practicing these steps consistently will significantly improve accuracy and speed in NMAT physics.<\/p>\n<h2>Key Takeaways for NMAT Physics<\/h2>\n<p>Momentum and collisions form a high-yield topic in NMAT Physics. Mastery of conservation laws, impulse,<br \/>\ncollision types, and vector reasoning provides a strong advantage in solving exam questions.<\/p>\n<p>With regular practice and conceptual clarity, these problems become systematic and predictable, making them<br \/>\none of the most scoring areas in the NMAT Physics section.<\/p>\n<h2>Momentum and Collisions: Problem Sets with Answer Keys (NMAT Physics)<\/h2>\n<h2>Problem Set 1: Momentum Basics and Impulse<\/h2>\n<ol>\n<li>A 2.0 kg cart moves to the right at 5.0 m\/s. What is its momentum?<\/li>\n<li>A 0.50 kg ball moving at 12 m\/s is brought to rest in 0.20 s. Find:<br \/>\n(a) the change in momentum, and (b) the average force applied.<\/li>\n<li>A 1.5 kg object initially at rest experiences an impulse of 9.0 N\u00b7s to the left.<br \/>\nWhat is its final velocity?<\/li>\n<li>A 1000 kg car moving at 20 m\/s applies brakes and stops in 5.0 s.<br \/>\nWhat is the average braking force (magnitude)?<\/li>\n<li>A force of 40 N acts on a body for 0.15 s. If the body\u2019s initial momentum is 6.0 kg\u00b7m\/s to the right<br \/>\nand the force acts to the left, what is the final momentum?<\/li>\n<\/ol>\n<h2>Answer Key 1<\/h2>\n<ol>\n<li><strong>p = mv = (2.0)(5.0) = 10 kg\u00b7m\/s<\/strong> (to the right)<\/li>\n<li>(a) <strong>\u0394p = m(vf \u2212 vi) = 0.50(0 \u2212 12) = \u22126.0 kg\u00b7m\/s<\/strong> (opposite the initial direction)<br \/>\n(b) <strong>Favg = \u0394p\/\u0394t = (\u22126.0)\/(0.20) = \u221230 N<\/strong> (30 N opposite the motion)<\/li>\n<li><strong>J = \u0394p = mv<\/strong> (since initial p = 0).<br \/>\n<strong>v = J\/m = (\u22129.0)\/1.5 = \u22126.0 m\/s<\/strong> (6.0 m\/s to the left)<\/li>\n<li><strong>\u0394p = m(0 \u2212 20) = 1000(\u221220) = \u221220000 kg\u00b7m\/s<\/strong><br \/>\n<strong>Favg = \u0394p\/\u0394t = (\u221220000)\/5.0 = \u22124000 N<\/strong><br \/>\nMagnitude: <strong>4000 N<\/strong><\/li>\n<li><strong>J = F\u0394t = (\u221240)(0.15) = \u22126.0 N\u00b7s<\/strong><br \/>\n<strong>pf = pi + J = 6.0 + (\u22126.0) = 0 kg\u00b7m\/s<\/strong><\/li>\n<\/ol>\n<hr \/>\n<h2>Problem Set 2: One-Dimensional Collisions (Momentum Conservation)<\/h2>\n<ol>\n<li>A 3.0 kg object moving at 4.0 m\/s to the right collides head-on with a 1.0 kg object moving at 2.0 m\/s<br \/>\nto the left. After collision, the 3.0 kg object moves at 2.0 m\/s to the right.<br \/>\nFind the final velocity of the 1.0 kg object.<\/li>\n<li>A 0.20 kg puck moving at 6.0 m\/s to the right collides with a stationary 0.30 kg puck.<br \/>\nAfter collision, the 0.20 kg puck moves at 1.5 m\/s to the right. Find the final speed of the 0.30 kg puck.<\/li>\n<li>Two carts move on a frictionless track. Cart A (2.0 kg) moves right at 3.0 m\/s,<br \/>\ncart B (4.0 kg) moves right at 1.0 m\/s. They collide and stick together.<br \/>\nWhat is their common velocity after collision?<\/li>\n<li>A 5.0 kg block at rest explodes into two pieces. One piece of mass 2.0 kg flies to the right at 6.0 m\/s.<br \/>\nWhat is the velocity of the other piece?<\/li>\n<li>A 1.0 kg ball moving at 8.0 m\/s to the right collides with a stationary 1.0 kg ball.<br \/>\nIf the collision is perfectly elastic, what are the final velocities of the two balls?<\/li>\n<\/ol>\n<h2>Answer Key 2<\/h2>\n<ol>\n<li>Take right as positive.<br \/>\nInitial momentum: <strong>p_i = 3(4) + 1(\u22122) = 12 \u2212 2 = 10<\/strong><br \/>\nFinal momentum: <strong>p_f = 3(2) + 1(v2) = 6 + v2<\/strong><br \/>\nSet equal: <strong>10 = 6 + v2 \u2192 v2 = 4.0 m\/s<\/strong> (to the right)<\/li>\n<li><strong>p_i = 0.20(6.0) + 0.30(0) = 1.2<\/strong><br \/>\n<strong>p_f = 0.20(1.5) + 0.30(v) = 0.30 + 0.30v<\/strong><br \/>\n<strong>1.2 = 0.30 + 0.30v \u2192 0.90 = 0.30v \u2192 v = 3.0 m\/s<\/strong><\/li>\n<li>Perfectly inelastic: stick together.<br \/>\n<strong>p_i = 2(3) + 4(1) = 6 + 4 = 10<\/strong><br \/>\nTotal mass: <strong>2 + 4 = 6<\/strong><br \/>\n<strong>v = p_i\/(m_total) = 10\/6 = 1.67 m\/s<\/strong> (to the right)<\/li>\n<li>Initial momentum is zero (at rest before explosion).<br \/>\nLet the second piece have mass <strong>3.0 kg<\/strong> and velocity <strong>v<\/strong>.<br \/>\n<strong>0 = 2(6) + 3(v) \u2192 0 = 12 + 3v \u2192 v = \u22124.0 m\/s<\/strong><br \/>\nSo it moves at <strong>4.0 m\/s to the left<\/strong>.<\/li>\n<li>In a 1D perfectly elastic collision between equal masses where one is initially at rest,<br \/>\nthe moving object stops and the stationary object takes its speed.<br \/>\nFinal velocities: <strong>v1 = 0 m\/s<\/strong>, <strong>v2 = 8.0 m\/s<\/strong> (to the right)<\/li>\n<\/ol>\n<hr \/>\n<h2>Problem Set 3: Perfectly Inelastic Collisions (Stick Together)<\/h2>\n<ol>\n<li>A 1.0 kg cart moving at 7.0 m\/s collides with a 2.0 kg cart moving at 1.0 m\/s in the same direction.<br \/>\nThey stick together. Find their final velocity.<\/li>\n<li>A 0.40 kg lump of clay moving at 5.0 m\/s hits and sticks to a 0.60 kg block at rest on a frictionless surface.<br \/>\nWhat is the final velocity of the combined mass?<\/li>\n<li>A 1200 kg car moving at 18 m\/s rear-ends a 1000 kg car moving at 10 m\/s. They lock together after impact.<br \/>\nFind their speed immediately after collision.<\/li>\n<li>A 2.0 kg object moving at 3.0 m\/s to the right sticks to a 1.0 kg object moving at 2.0 m\/s to the left.<br \/>\nFind the final velocity and its direction.<\/li>\n<li>A bullet of mass 0.010 kg traveling at 400 m\/s embeds in a 2.0 kg wooden block initially at rest.<br \/>\nWhat is the block+bullet speed right after the collision?<\/li>\n<\/ol>\n<h2>Answer Key 3<\/h2>\n<ol>\n<li><strong>p_i = 1(7) + 2(1) = 7 + 2 = 9<\/strong><br \/>\nTotal mass = <strong>3<\/strong><br \/>\n<strong>v = 9\/3 = 3.0 m\/s<\/strong> (same direction)<\/li>\n<li><strong>p_i = 0.40(5.0) + 0 = 2.0<\/strong><br \/>\nTotal mass = <strong>1.0<\/strong><br \/>\n<strong>v = 2.0\/1.0 = 2.0 m\/s<\/strong><\/li>\n<li><strong>p_i = 1200(18) + 1000(10) = 21600 + 10000 = 31600<\/strong><br \/>\nTotal mass = <strong>2200<\/strong><br \/>\n<strong>v = 31600\/2200 = 14.36 m\/s<\/strong><\/li>\n<li>Take right as positive:<br \/>\n<strong>p_i = 2(3) + 1(\u22122) = 6 \u2212 2 = 4<\/strong><br \/>\nTotal mass = <strong>3<\/strong><br \/>\n<strong>v = 4\/3 = 1.33 m\/s<\/strong> (to the right)<\/li>\n<li><strong>p_i = 0.010(400) = 4.0<\/strong><br \/>\nTotal mass = <strong>2.010 kg<\/strong><br \/>\n<strong>v = 4.0\/2.010 = 1.99 m\/s<\/strong> (approximately)<\/li>\n<\/ol>\n<hr \/>\n<h2>Problem Set 4: Kinetic Energy and Collision Concepts<\/h2>\n<ol>\n<li>A 2.0 kg ball moving at 6.0 m\/s has what kinetic energy?<\/li>\n<li>Two 1.0 kg carts collide and stick together. One cart moves at 4.0 m\/s, the other is at rest.<br \/>\nFind the kinetic energy before and after collision, then determine how much kinetic energy is lost.<\/li>\n<li>Which type of collision results in the maximum loss of kinetic energy:<br \/>\nelastic, inelastic, or perfectly inelastic?<\/li>\n<li>A 0.50 kg object moving at 10 m\/s undergoes a perfectly elastic collision and rebounds with the same speed.<br \/>\nCompare the magnitude of momentum before and after.<\/li>\n<li>A truck and a car collide. Total momentum is conserved but kinetic energy decreases.<br \/>\nWhat type of collision is this?<\/li>\n<\/ol>\n<h2>Answer Key 4<\/h2>\n<ol>\n<li><strong>K = (1\/2)mv<sup>2<\/sup> = (1\/2)(2.0)(6.0<sup>2<\/sup>) = 36 J<\/strong><\/li>\n<li>Initial KE: <strong>K_i = (1\/2)(1)(4<sup>2<\/sup>) = 8 J<\/strong><br \/>\nFinal velocity: <strong>v = (1\u00b74 + 1\u00b70)\/(2) = 2 m\/s<\/strong><br \/>\nFinal KE: <strong>K_f = (1\/2)(2)(2<sup>2<\/sup>) = 4 J<\/strong><br \/>\nKE lost: <strong>8 \u2212 4 = 4 J<\/strong><\/li>\n<li><strong>Perfectly inelastic collision<\/strong><\/li>\n<li>Momentum magnitude before: <strong>p = mv = 0.50(10) = 5.0 kg\u00b7m\/s<\/strong><br \/>\nAfter rebound, speed is the same, so momentum magnitude is still <strong>5.0 kg\u00b7m\/s<\/strong><br \/>\n(direction changes, but magnitude is unchanged).<\/li>\n<li><strong>Inelastic collision<\/strong><\/li>\n<\/ol>\n<hr \/>\n<h2>Problem Set 5: Mixed NMAT-Style Word Problems<\/h2>\n<ol>\n<li>A 0.25 kg ball is hit, changing its velocity from 2.0 m\/s to 14 m\/s in the same direction.<br \/>\nIf contact time is 0.050 s, find the average force exerted on the ball.<\/li>\n<li>A 60 kg skater initially at rest pushes off a 40 kg skater. After pushing, the 60 kg skater moves left at 2.0 m\/s.<br \/>\nFind the velocity of the 40 kg skater.<\/li>\n<li>A 2.0 kg cart moving right at 5.0 m\/s collides elastically with a 1.0 kg cart at rest.<br \/>\nAfter collision, the 2.0 kg cart moves at 1.0 m\/s to the right.<br \/>\nFind the final velocity of the 1.0 kg cart (use momentum conservation).<\/li>\n<li>A 0.020 kg bullet moving at 300 m\/s passes through a 1.0 kg block initially at rest and exits at 100 m\/s<br \/>\n(same direction). Find the block\u2019s speed immediately after the bullet exits (ignore friction).<\/li>\n<li>A 0.50 kg ball moving at 8.0 m\/s to the right collides head-on with a 0.50 kg ball moving at 4.0 m\/s to the left.<br \/>\nIf they stick together, find their velocity after collision.<\/li>\n<\/ol>\n<h2>Answer Key 5<\/h2>\n<ol>\n<li><strong>\u0394p = m(vf \u2212 vi) = 0.25(14 \u2212 2) = 0.25(12) = 3.0 kg\u00b7m\/s<\/strong><br \/>\n<strong>Favg = \u0394p\/\u0394t = 3.0\/0.050 = 60 N<\/strong><\/li>\n<li>Initial momentum = 0.<br \/>\nTake right as positive; left is negative.<br \/>\n<strong>0 = 60(\u22122.0) + 40(v)<\/strong><br \/>\n<strong>0 = \u2212120 + 40v \u2192 v = 3.0 m\/s<\/strong> (to the right)<\/li>\n<li><strong>p_i = 2(5) + 1(0) = 10<\/strong><br \/>\n<strong>p_f = 2(1) + 1(v) = 2 + v<\/strong><br \/>\n<strong>10 = 2 + v \u2192 v = 8.0 m\/s<\/strong> (to the right)<\/li>\n<li>Momentum conservation (block + bullet):<br \/>\n<strong>p_i = 0.020(300) = 6.0<\/strong><br \/>\nFinal momentum: <strong>p_f = 0.020(100) + 1.0(v) = 2.0 + v<\/strong><br \/>\n<strong>6.0 = 2.0 + v \u2192 v = 4.0 m\/s<\/strong> (block to the right)<\/li>\n<li>Perfectly inelastic; stick together. Take right positive.<br \/>\n<strong>p_i = 0.50(8.0) + 0.50(\u22124.0) = 4.0 \u2212 2.0 = 2.0<\/strong><br \/>\nTotal mass = <strong>1.0 kg<\/strong><br \/>\n<strong>v = 2.0\/1.0 = 2.0 m\/s<\/strong> (to the right)<\/li>\n<\/ol>\n<blockquote class=\"wp-embedded-content\" data-secret=\"bCc22oKM7B\"><p><a href=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-physics-review.html\">NMAT Physics Review: NMAT Study Guide<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; visibility: hidden;\" title=\"&#8220;NMAT Physics Review: NMAT Study Guide&#8221; &#8212; Study English at 3D ACADEMY, a Language School in Cebu, Philippines\" src=\"https:\/\/3d-universal.com\/en\/blogs\/nmat-physics-review.html\/embed#?secret=pvTNUBVEN7#?secret=bCc22oKM7B\" data-secret=\"bCc22oKM7B\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"rQUz6hECGp\"><p><a href=\"https:\/\/3d-universal.com\/en\/philippines-university-guide\/nmat-study-guide\">NMAT Study Guide: Complete Preparation Guide for Medical School in the Philippines<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; visibility: hidden;\" title=\"&#8220;NMAT Study Guide: Complete Preparation Guide for Medical School in the Philippines&#8221; &#8212; Study English at 3D ACADEMY, a Language School in Cebu, Philippines\" src=\"https:\/\/3d-universal.com\/en\/philippines-university-guide\/nmat-study-guide\/embed#?secret=dmN17lMjq9#?secret=rQUz6hECGp\" data-secret=\"rQUz6hECGp\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":21109,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_kad_blocks_custom_css":"","_kad_blocks_head_custom_js":"","_kad_blocks_body_custom_js":"","_kad_blocks_footer_custom_js":"","footnotes":""},"categories":[117],"tags":[],"class_list":["post-21108","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nmat-physics-review"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v25.6 (Yoast SEO v25.6) - 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