A 41.2-kg boy, riding a 2.57-kg skateboard at a velocity of +5.82 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.12 m/s, 9.72° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Question

A 41.2-kg boy, riding a 2.57-kg skateboard at a velocity of +5.82 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is 6.12 m/s, 9.72° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Answer 1

Read your question carefully. It contains the answer.

"..the boy's velocity relative to the sidewalk is 6.12 m/s, 9.72° above the horizontal....Just after leaving contact with the board"

Answer 2

To find the skateboard's velocity relative to the sidewalk at the instant the boy jumps, we need to use the law of conservation of momentum.

The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the boy and the skateboard form a closed system.

The momentum of an object is given by the product of its mass and velocity. So, we can express the momentum of the boy before and after he jumps as follows:

Momentum before jump = (mass of boy) x (velocity of boy before jump)
Momentum after jump = (mass of boy) x (velocity of boy after jump) + (mass of skateboard) x (velocity of skateboard after jump)

According to the law of conservation of momentum, the momentum before the jump should be equal to the momentum after the jump.

Let's use this information to solve the problem:

Given:
Mass of boy (mb) = 41.2 kg
Mass of skateboard (ms) = 2.57 kg
Velocity of boy before jump (vb,before) = +5.82 m/s
Velocity of boy after jump (vb,after) = 6.12 m/s
Angle above horizontal (θ) = 9.72°

Momentum before jump = Momentum after jump

(mb)(vb,before) = (mb)(vb,after) + (ms)(vs,after)

Plugging in the given values:
(41.2 kg)(5.82 m/s) = (41.2 kg)(6.12 m/s) + (2.57 kg)(vs,after)

Now we solve for the velocity of the skateboard after the boy jumps:

240.024 kg·m/s = 252.744 kg·m/s + (2.57 kg)(vs,after)

Subtracting 252.744 kg·m/s from both sides:

-12.72 kg·m/s = (2.57 kg)(vs,after)

Dividing both sides by 2.57 kg:

vs,after = -12.72 kg·m/s / 2.57 kg ≈ -4.95 m/s

Therefore, the skateboard's velocity relative to the sidewalk at the instant the boy jumps is approximately -4.95 m/s (the negative sign indicates the direction is opposite to the initial direction of the boy's velocity).