Asmall, 0.10 kg cart is moving at 1.54 m/s when it collides with a larger 1.52 kg cart at rest. after the elastic collision, the small cart recoils (i. e. bounces off the larger cart and travels in the opposite direction) at a speed of 0.76 m/s.(a) what is the magnitude (absolute value) of the change in momentum for the small cart? kg m/s(b) what is the speed (absolute value of the velocity) of the larger cart after the collision? m/s

a. the absolute value of the change in the momentum of the small car is 0.078  

b. the velocity of the larger car after the collision is 0.1513 m/s

Explanation:

The linear momentum P is calculated as:

P = MV

Where M is the mass and V the velocity

Therefore, for calculated the change of the linear momentum of the small cart, we get:

ΔP

where in the inicial momentum and is the final momentum of the small cart. Replacing the values, we get:

0.10 kg (0.76) -0.10(1.54) = -0.078 kg m/s

The absolute value: 0.078 kg m/s

On the other hand, using the law of the conservation of linear momentum, we get:

Where is the linear momentum of the sistem before the collision and is the linear momentum after the collision.

Where is the linear momentum of the small cart after the collision and is the linear momentum of the larger cart after the collision

so:

(0.10 kg)(1.54 m/s) = (0.10 kg)(-0.76 m/s) + (1.52 kg)(V)

Note: we choose the first direction of the small car as positive.

Solving for V:

V = 0.1513 m/s

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.