Question

A small, 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

Answer 1

Answer:

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:

$P_{fc}-P_{ic} =$ΔP

where $P_{ic}$ in the inicial momentum and $P_{fc}$ 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:

$P_i = P_f$

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

$P_i=P_{ic}\\P_f=P_{fc} + P_{ft}$

Where $P_{fc}$ is the linear momentum of the small cart after the collision and $P_{ft}$ 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:

$\frac{0.10(1.54)+0.10(0.76)}{1.52} = V$

V = 0.1513 m/s