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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 4

Lesson 5: Conservation of linear momentum

# Bouncing fruit collision example

When an object with known velocity collides with an object at rest, and the velocity of one of the objects after the collision is known, conservation of momentum can be used to determine the velocity of the other object after the collision. It is important to consider the directions of velocity and momentum when interpreting signs in the calculation. Created by David SantoPietro.

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• I guess the apple overcame the string pulling it after it was hit by the orange .. doesn't that mean that a part of the momentum of the orange was used to overcome the string tension force .. so that less momentum will transfer to the movement of the apple .. !!

is that right ? •  We're assuming the stem was very weak and that the apple was not held tightly to the tree, otherwise yes, we would have to take into account the force from tension.
• If the orange hits the apple at the apex of its trajectory, then shouldn't its velocity be 0 m/s (since that's the point at which it changes direction)?? • What would happen if the two stuck together? • Does the conservation of momentum apply in the dodge ball example discussed in last video ? • How do you find the velocity of an object after a collision if they don't give you the mass of the object? • For a unidimensional collision between two bodies of masses M and m, each with a initial speed of V1 and v1, respectively. The speeds after the collision become V2 and v2.
Conservation of energy:
M(V1)² + m(v1)² = M(V2)² + m(v2)²
M[(V1)² - (V2)²] = m[(v2)² - (v1)²] 1
Conservation of linear momentum:
M(V1) + m(v1) = M(V2) + m(v2)
M[(V1) - (V2)] = m[(v2) - (v1)] 2
Since a² - b² = (a + b)(a - b) , dividing equation 1 by equation 2, we have
(V2) + (V1) = (v2) + (v1)
Which should probably be of some help
• How is momentum conserved when a ball hits the ground? • Momentum is conserved in a system when there are no external forces acting on the system. Keeping this in mind, we can analyze the problem from the perspectives of different systems.
First, how about just the ball. When the ball hits the wall, the wall provides a normal force that is external to the system, so its correct to say that momentum isn't conserved.
How about the ball plus the wall? Now the normal force from the wall to the ball is an internal force, and momentum should be conserved if there are no other forces. If you just had a huge wall floating in space, the ball would transfer some of its momentum to the wall (which would correspond to an almost nonexistent change in velocity because the wall is so massive). But if the wall is like a normal wall and is fixed to the ground, the ground will exert an external force on the wall against the wall's minuscule motion, making momentum again not conserved.
But if you really expand your system, then the ball transfers some of its momentum to the wall, which transfers it to the ground, which is still in the system.
• If I throw a 5 kg ball at a wall with a velocity of 20m/s , the momentum of the ball is 100kgm/s. If the ball comes back with the same speed but in the opposite direction with a momentum of -100 kgm/s, isn't the law of conservation of momentum defied, since the wall is stationary (has a momentum of 0) and the ball lost 200 kgm/s? • Why can we not use the conservation formula if there is external impulse? • Momentum is always conserved for any closed system.
Example if you have two billiard balls colliding, the momentum between the two balls remains unchanged before and after collision. If you push one of the balls, then the momentum of the two ball system will change since your hand is external to that system. Now if you change the system to include your body in addition to the two balls, the total momentum of the two balls and yourself will remain unchanged when you push a ball (assuming there are no other external forces such as friction). This example works best when you think about the two balls and yourself floating in space with no friction.
• How did an orange with total momentum of 2.0kgf managed to pass all of its momentum and then some more to move the apple with 2.1kgf?
And then even generate some more energy out of nowhere to also accelerate itself too; generating more work (10% more than all energy it had, 2.2kgf total from its 2.0).
If it could accelerate the apple as much as it like, as long as it also accelerate itself proportionally on the opposite direction, what is preventing it from accelerating the apple to 100kgf of momentum and itself to -98kgf? Or infinitely at opposite directions like those glitched videogames. • (Just a note, we generally use kg*m/s as the unit of momentum and kg*m^2/s^2 or Joules as the unit of energy.)

Momentum is conserved in this situation because right after the collision, the apple's momentum is +2.1 kg*m/s while the orange's momentum at that instant is -0.1 kg*m/s. We add these together to get +2.1 kg*m/s+(-0.1 kg*m/s)= +2.0 kg*m/s, which is the same as the system's momentum right before the collision. Basically the apple can have a greater momentum than the orange initially did because the orange has a lesser velocity after the collision than it did before.

If we use the velocities found with conservation of momentum in the video, we find that the system actually loses kinetic energy during the collision. Kinetic energy = 1/2 mv^2, so the orange's initial kinetic energy is (1/2)(0.4)(5)^2 = 5 Joules. Since the apple begins at rest, its initial kinetic energy is 0 J, so the entire system's initial kinetic energy is 5 J. Substituting the fruits' final velocities into the kinetic energy equation tells us that the orange's final kinetic energy is 0.0125 J, and the apple's final kinetic energy is 3.15 J. Adding these together, we get the system's final kinetic energy: 3.1625 J.

As long as the total final kinetic energy was less than or equal to the total initial kinetic energy, the fruits could have any velocities. If the collision released some stored energy (like in a spring maybe), the total final kinetic energy could even be greater than the initial. 