A 40.0-kg boy, riding a 2.50-kg skateboard at a velocity of +5.30 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 6.00 m/s, 9.50o 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.
7.32
To find the skateboard's velocity relative to the sidewalk at this instant, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces are acting on it. In this case, the system consists of the boy and the skateboard.
Initially, the boy and the skateboard are moving together with a velocity of +5.30 m/s across the sidewalk. We can calculate the total momentum of the system using the equation:
Total momentum = (mass of the boy × velocity of the boy) + (mass of the skateboard × velocity of the skateboard)
Total momentum = (40.0 kg × 5.30 m/s) + (2.50 kg × 5.30 m/s)
Total momentum = (212 kg·m/s) + (13.25 kg·m/s)
Total momentum = 225.25 kg·m/s
When the boy jumps forward to leap over the wall, he has a velocity of 6.00 m/s, 9.50° above the horizontal. To find the skateboard's velocity relative to the sidewalk at this instant, we need to find the horizontal component of the boy's velocity. We can use the trigonometric function cosine to do this.
Horizontal component of velocity = 6.00 m/s × cos(9.50°)
Horizontal component of velocity = 6.00 m/s × 0.976
Horizontal component of velocity ≈ 5.856 m/s
Since the skateboard and the boy were initially moving together, the horizontal component of the skateboard's velocity relative to the sidewalk remains the same.
Therefore, the skateboard's velocity relative to the sidewalk at this instant is approximately +5.856 m/s in the horizontal direction.