Answer:
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.
