Answer:
The speed of the two carts after the collision is 10 m/s.
Explanation:
Hi there!
The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:
initial momentum = mA · vA1 + mB · vB1
final momentum = (mA + mB) · vAB2
Where:
mA = mass of cart A = 0.500 kg
vA1 = velocity of cart A before the collision = 100 m/s
mB = mass of cart B = 1.50 kg.
vB1 = velocity of cart B before the collision = - 20 m/s
vAB2 = velocity of the carts that move as a single object = unknown.
(notice that we have considered leftward as negative direction)
Since the momentum of system remains constant:
initial momentum = final momentum
mA · vA1 + mB · vB1 = (mA + mB) · vAB2
Solving for vAB2:
(mA · vA1 + mB · vB1) / (mA + mB) = vAB2
(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2
vAB2 = 10 m/s
The speed of the two carts after the collision is 10 m/s.
