A20-kg crate is sitting on a frictionless ice rink. a child standing at the side of the ice rink uses a slingshot to launch a 1.5-kg beanbag at the crate. assume that the beanbag strikes the crate horizontally in the +x direction with a speed of 10 m/s. the beanbag bounces straight back in the -x direction with a speed of 6 m/s. using conservation of momentum, calculate the speed of the crate after the beanbag hits it.

The speed of the crate after the beanbag hits it is 1.2 m/s.

Explanation:

Hi there!

The momentum of the system beanbag-crate remains the same after the collision, i.e., the momentum of the system is conserved. The momentum of the system is calculated by adding the momenta of each object. So, the initial momentum (before the collision) is calculated as follows:

initial momentum = momentum of the crate + momentum of the beanbag

initial momentum = mc · vc + mb · vb

Where:

mc = mass of the crate.

vc = initial velocity of the crate.

mb = mass of the beanbag

vb = initial mass of the beanbag

With the data we have, we can calculate the initial momentum:

initial momentum  = 20 kg · 0 m/s + 1.5 kg · 10 m/s = 15 kg · m/s

Now, let´s write the equation of the momentum of the system after the collision:

final momentum = mc · vc´ + mb · vb´

Where vc´ and vb´ are the final velocity of the crate and the beanbag respectively. Let´s replace with the data we have:

final momentum = 20 kg · vc´ + 1.5 · (-6 m/s)

Since

initial momentum = final momentum

Then:

15 kg · m/s = 20 kg · vc´ + 1.5 kg · (-6 m/s)

Solving for vc´:

(15 kg · m/s + 9 kg · m/s) / 20 kg = vc´

vc´ = 1.2 m/s

The speed of the crate after the beanbag hits it is 1.2 m/s.

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