Question

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the person is 68.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 21.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard, relative to the ground.

Answer 1

The recoil speed of the person and the skateboard, relative to the ground, is 0.671 m/s in the opposite direction of the thrown brick.

Step-by-step explanation:

To determine the recoil speed of the person and the skateboard after throwing the brick, we can use the principle of conservation of momentum.

The initial momentum of the system (person, skateboard, and brick) is equal to the final momentum of the system.

Let's assume the recoil speed of the person and the skateboard after throwing the brick is v, and the velocity of the brick relative to the ground is 21.0 m/s.

Initial momentum = (person's mass + skateboard's mass + brick's mass) * (0 m/s) = (68.0 kg + 4.10 kg + 2.50 kg) * (0 m/s) = 0 kg*m/s

Final momentum = (person's mass + skateboard's mass + brick's mass) * (recoil speed) + (brick's mass) * (velocity relative to the ground)

Final momentum = (68.0 kg + 4.10 kg + 2.50 kg) * v + (2.50 kg) * (21.0 m/s)

Equating the initial and final momenta, we can solve for v:

0 = (74.6 kg) * v + (2.50 kg) * (21.0 m/s)

Simplifying the equation, we have:

(74.6 kg) * v = -(2.50 kg) * (21.0 m/s)

v = -((2.50 kg) * (21.0 m/s)) / (74.6 kg)

v = -0.671 m/s

Therefore, the recoil speed of the person and the skateboard, relative to the ground, is 0.671 m/s in the opposite direction of the thrown brick.