{"id":1777,"date":"2018-04-10T05:21:52","date_gmt":"2018-04-09T21:21:52","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1777"},"modified":"2023-08-09T23:51:49","modified_gmt":"2023-08-09T23:51:49","slug":"impulse-momentum-collisions-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/impulse-momentum-collisions-problems-and-solutions.htm","title":{"rendered":"Impulse Momentum Collisions \u2013 Problems and Solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Impulse Momentum Collisions \u2013 Problems and Solutions<\/strong><\/span><\/p>\n

1. A 40-gram rubber bullet shot horizontally to the wall, as shown in the figure below. The bullet is reflected at the same speed. What is the change in momentum<\/a> of the ball?<\/span><\/p>\n

Known :\"Impulse,<\/u><\/span><\/p>\n

Mass<\/a> (m) = 40 gram = 0.04 kg<\/span><\/p>\n

Initial velocity (vo<\/sub>) = – 50 m\/s<\/span><\/p>\n

Final velocity (vt<\/sub>) = 50 m\/s<\/span><\/p>\n

Direction of ball’s displacement (direction of velocity) is opposite so that the initial velocity and the final velocity have opposite signs.<\/span><\/p>\n

Wanted :<\/u> The change in momentum of ball<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The change in momentum :<\/span><\/p>\n

\u0394p = m (vt<\/sub> \u2013 vo<\/sub>) = (0.04)(50 \u2013 (-50)) = (0.04)(50 + 50)<\/span><\/p>\n

\u0394p = (0.04)(100) = 4 N.s<\/span><\/p>\n

2. A 75-gram rubber ball thrown horizontally to hit the wall, as shown in figure below. The ball is reflected at the same speed. What is the impulse.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball (m) = 75 gram = 0.075 kg\"Impulse,<\/span><\/p>\n

Initial velocity (vo<\/sub>) = -20 m\/s<\/span><\/p>\n

Final velocity (vt<\/sub>) = 20 m\/s <\/span><\/p>\n

Direction of ball’s displacement (direction of velocity) is opposite so that the initial velocity and the final velocity have opposite signs.<\/span><\/p>\n

Wanted :<\/u> Impulse<\/a><\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Impulse = The change in momentum<\/span><\/p>\n

I = \u0394p<\/span><\/p>\n

I = m (vt<\/sub> \u2013 vo<\/sub>) = 0.075 (20 \u2013 (-20)) = 0.075 (20 + 20) = 0.075 (40)<\/span><\/p>\n

I = 3 N s = 3 Newton second<\/span><\/p>\n

Perfectly elastic collision<\/b><\/a><\/span><\/p>\n

3. Two balls, A and B, approach each other along a horizontal plane. Velocity of ball A, vA<\/sub> = 4 m\/s and velocity of ball B, vB<\/sub> = 6 m\/s. The collision is perfectly elastic. The velocity of object B after collision is 4 m\/s. What is the velocity of ball A after collision.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball A (mA<\/sub>) = m\"Impulse,<\/span><\/p>\n

Mass of ball B (mB<\/sub>) = m<\/span><\/p>\n

Velocity of ball A before collision (vA<\/sub>) = 4 m\/s<\/span><\/p>\n

Velocity of ball B before collision (vB<\/sub>) = 6 m\/s <\/span><\/p>\n

Velocity of ball B after collision (vB<\/sub>\u2019) = 4 m\/s<\/span><\/p>\n

Wanted:<\/u> Velocity of ball A after a collision (vA<\/sub>\u2019)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Mass of balls are same, so that :<\/span><\/p>\n

Velocity of ball A before collision (vA<\/sub>) = velocity of ball B after collision (vB<\/sub>\u2019) = 4 m\/s <\/span><\/p>\n

Velocity of ball B before collision (vB<\/sub>) = velocity of ball A after collision (vA<\/sub>\u2019) = 6 m\/s <\/span><\/p>\n

Partially Elastic Collisions<\/b><\/a><\/span><\/p>\n

4. Two objects <\/span>with <\/span>the same mass move in a straight line approaching <\/span>each other, as shown in the figure below. <\/span>If v<\/span>2 <\/span><\/sub>is the velocity of the object <\/span>2, <\/span>after the collision to the right at 5 m.s<\/span>-1<\/span><\/sup>, then the velocity <\/span>of the object <\/span>(1) after the collision is…<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Object’s mass = m\"Impulse,<\/span><\/p>\n

Velocity of object 1 before collision (v1<\/sub>) = 8 m\/s<\/span><\/p>\n

Velocity of object 2 before collision (v2<\/sub>) = 10 m\/s <\/span><\/p>\n

Velocity of object 2 after collision (v2<\/sub>‘) = 5 m\/s<\/span><\/p>\n

Wanted :<\/u> Velocity of object 1 after collision (v1<\/sub>‘) <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

m1<\/sub> v1<\/sub>+ m2<\/sub> v2<\/sub> = m1<\/sub> v1<\/sub>\u2019 + m2<\/sub> v2<\/sub>\u2019<\/span><\/p>\n

m (v1<\/sub> + v2<\/sub>) = m (v1<\/sub>\u2019 + v2<\/sub>\u2019)<\/span><\/p>\n

v1<\/sub> + v2<\/sub> = v1<\/sub>\u2019 + v2<\/sub>\u2019<\/span><\/p>\n

8 + 10 = v1<\/sub>\u2019 + 5<\/span><\/p>\n

18 = v1<\/sub>\u2019 + 5<\/span><\/p>\n

v1<\/sub>\u2019 = 18-5<\/span><\/p>\n

v1<\/sub>\u2019 = 13 m\/s<\/span><\/p>\n

5. Object A with mass of 4 kg and object B with mass of 5 kg approach each other along a horizontal plane as shown in figure below. After collision, velocity of ball A = 4 m\/s and velocity of ball B = 2 m\/s. What is the velocity of ball B before collision.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of object A (mA<\/sub>) = 4 kg\"Impulse,<\/span><\/p>\n

Mass of object B B (mB<\/sub>) = 5 kg<\/span><\/p>\n

Velocity of object A before collision (vA<\/sub>) = 6 m\/s<\/span><\/p>\n

Velocity of object A after collision (vA<\/sub>\u2019) = 4 m\/s<\/span><\/p>\n

Velocity of object B after collision (vB<\/sub>\u2019) = -2 m\/s<\/span><\/p>\n

Plus and minus sign indicates that balls have opposite direction.<\/span><\/p>\n

Wanted :<\/u> Velocity of ball B before collision (vB<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

mA<\/sub> vA <\/sub>+ mB<\/sub> vB<\/sub> = mA<\/sub> vA<\/sub>\u2019 + mB<\/sub> vB<\/sub>\u2019<\/span><\/p>\n

(4)(6) + (5)(-2) = (4)(4) + (5)(vB<\/sub>\u2019)<\/span><\/p>\n

24 \u2013 10 = 16 + 5(vB<\/sub>\u2019) <\/span><\/p>\n

14 – 16 = 5 (vB<\/sub>\u2019) <\/span><\/p>\n

-2 = 5 (vB<\/sub>\u2019) <\/span><\/p>\n

vB<\/sub>\u2019 = -2\/5 <\/span><\/p>\n

vB<\/sub>\u2019 = -0.4<\/span><\/p>\n

Minus sign indicates that direction of ball after collision is opposite with the direction of ball before collision.<\/span><\/p>\n

I<\/span>ne<\/span>lastic Collisions<\/span><\/b><\/a><\/span><\/p>\n

6. A 2-kg ball and a 1-kg ball approach each other along a horizontal plane, as shown in figure below.<\/span><\/p>\n

Velocity of ball 1, v1<\/sub> = 2 ms-1 <\/sup>and velocity of ball 2, v2<\/sub> = 4 ms-1<\/sup>. After the collision, both stick together. What is the velocity after the collision?<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball 1 (m1<\/sub>) = 2 kg\"Impulse,<\/span><\/p>\n

Mass of ball 2 (m2<\/sub>) = 1 kg<\/span><\/p>\n

Velocity of ball 1 before collision (v1<\/sub>) = 2 m\/s<\/span><\/p>\n

Velocity of ball 2 before collision (v2<\/sub>) = -4 m\/s <\/span><\/p>\n

Direction of ball’s displacement<\/a> (direction of velocity) is opposite so that velocity of ball 1 and velocity of ball 2 have opposite signs. <\/span><\/p>\n

Wanted :<\/u> Velocity after collision (v\u2019) <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

m1<\/sub> v1<\/sub> + m2 <\/sub>v2<\/sub> = (m1 <\/sub>+ m2<\/sub>) v\u2019<\/span><\/p>\n

(2)(2) + (1)(-4) = (2 + 1) v\u2019<\/span><\/p>\n

4 \u2013 4 = (3) v\u2019<\/span><\/p>\n

0 = (3) v\u2019<\/span><\/p>\n

v\u2019 = 0 <\/span><\/p>\n

1. What is the impulse-momentum theorem?<\/strong><\/span><\/p>\n

Answer:<\/strong> The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. Mathematically, I<\/span>m<\/span>p<\/span>u<\/span>l<\/span>se<\/span>=<\/span><\/span>F<\/span>\u00d7<\/span><\/span>\u0394<\/span>t<\/span>=<\/span><\/span>\u0394<\/span>p<\/span><\/span><\/span><\/span><\/span>, where F<\/span><\/span><\/span><\/span><\/span> is the average force applied, \u0394<\/span>t<\/span><\/span><\/span><\/span><\/span> is the time for which the force is applied, and \u0394<\/span>p<\/span><\/span><\/span><\/span><\/span> is the change in momentum.<\/span><\/p>\n

2. How is the conservation of momentum principle applied during collisions?<\/strong><\/span><\/p>\n

Answer:<\/strong> In an isolated system (no external forces), the total momentum before a collision is equal to the total momentum after the collision. This is true for both elastic and inelastic collisions.<\/span><\/p>\n

3. Differentiate between an elastic and an inelastic collision.<\/strong><\/span><\/p>\n

Answer:<\/strong> In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an inelastic collision, while momentum is conserved, kinetic energy is not necessarily conserved, and some may be transformed into other forms of energy like heat or sound.<\/span><\/p>\n

4. Why does a bouncy ball eventually stop bouncing when dropped from a certain height?<\/strong><\/span><\/p>\n

Answer:<\/strong> Even though the ball undergoes approximately elastic collisions with the ground, not all of its kinetic energy is converted back into potential energy. Some energy is lost to other forms such as heat, sound, and internal deformations in the ball. As a result, with each bounce, the height decreases until the ball eventually stops.<\/span><\/p>\n

5. How does impulse relate to a change in velocity?<\/strong><\/span><\/p>\n

Answer:<\/strong> Impulse is the product of average force and the time for which it acts on an object. Given the impulse-momentum theorem, impulse is equal to the change in momentum. For an object of constant mass, momentum (p<\/span><\/span><\/span><\/span><\/span>) is ma<\/span>ss<\/span>(<\/span>m<\/span>)<\/span>\u00d7<\/span><\/span>v<\/span>e<\/span>l<\/span>oc<\/span>i<\/span>t<\/span>y<\/span>(<\/span>v<\/span>)<\/span><\/span><\/span><\/span><\/span>. Hence, a change in momentum translates to a change in velocity.<\/span><\/p>\n

6. In a two-car collision where both cars come to a complete stop, is momentum still conserved?<\/strong><\/span><\/p>\n

Answer:<\/strong> Yes, the momentum is still conserved. The total momentum of the two cars before the collision would be equal to the total momentum after the collision. Since both cars come to a stop, their individual momenta are zero, but their combined momentum is still the same as before the collision.<\/span><\/p>\n

7. How does the impulse delivered by an airbag in a car reduce the impact on a passenger during a collision?<\/strong><\/span><\/p>\n

Answer:<\/strong> An airbag increases the time over which the deceleration of the passenger occurs. By extending the time, the average force experienced by the passenger (due to the impulse-momentum theorem) is reduced, thus lessening the injury risk.<\/span><\/p>\n

8. Why do athletes perform a “roll” when landing from a significant height?<\/strong><\/span><\/p>\n

Answer:<\/strong> By rolling upon landing, an athlete increases the duration of the impact. This longer time decreases the average force experienced during the landing, reducing the risk of injury. It’s an application of the impulse-momentum theorem in real life.<\/span><\/p>\n

9. How is momentum related to Newton’s third law of motion?<\/strong><\/span><\/p>\n

Answer:<\/strong> Newton’s third law states that for every action, there is an equal and opposite reaction. In collisions, one object exerts a force on the second, and the second exerts an equal and opposite force on the first. This means the change in momentum (or impulse) experienced by one object is equal in magnitude and opposite in direction to that of the other object, ensuring the conservation of momentum.<\/span><\/p>\n

10. How does the angle of incidence in a collision affect the resultant velocities of colliding objects?<\/strong><\/span><\/p>\n

Answer:<\/strong> The angle of incidence can determine the direction of the resultant velocities post-collision. In two-dimensional collisions, the conservation of momentum must be applied separately for the horizontal and vertical components. Depending on the angle of collision and the properties of the objects, this can lead to various outcomes in terms of direction and magnitude of the resultant velocities.<\/span><\/p>\n

    \n
  1. What is the relationship between impulse and momentum?<\/strong><\/span>\n