Question

A 52 kg cart is moving across a horizontal floor at 2.0 m/s toward the north. A 74 kg boy riding in the cart jumps backward off the cart so that he hits the floor with zero velocity. What impulse (magnitude and direction) does the cart give to the boy?

Answer 1

The cart imparts an impulse of 104 N·s southward to the boy.

Step-by-step explanation:

When the boy jumps backward off the moving cart, conservation of linear momentum can be applied to determine the impulse imparted by the cart to the boy. The initial momentum of the system (cart and boy) is given by the product of mass and velocity. Before the jump, the momentum is (52 kg + 74 kg) * 2.0 m/s = 252 kg·m/s northward. After the jump, the boy hits the floor with zero velocity, so the final momentum is (52 kg) * v (velocity of the cart) southward.

Using the principle of conservation of linear momentum:

`\[ \text{Initial momentum} = \text{Final momentum} \]\[ (52 \, \text{kg} + 74 \, \text{kg}) * 2.0 \, \text{m/s} = (52 \, \text{kg}) * v \]\[ v = \frac{(52 \, \text{kg} + 74 \, \text{kg}) * 2.0 \, \text{m/s}}{52 \, \text{kg}} = 4.0 \, \text{m/s} \]`

The change in velocity (Delta v ) is the final velocity (4.0 m/s) minus the initial velocity (2.0 m/s), which is 2.0 m/s southward. Now, to find the impulse (text{Impulse} =Delta p )), multiply the mass of the boy by ( Delta v ):

`\[ \text{Impulse} = 74 \, \text{kg} * 2.0 \, \text{m/s} = 148 \, \text{N·s} \]`

Therefore, the cart imparts an impulse of 104 N·s southward to the boy.