A 39.2-kg boy, riding a 2.39-kg skateboard at a velocity of +4.97 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 6.48 m/s, 9.54° 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?
V = 6.48*cos9.54 = 6.39 m/s.
the above answer is wrong. don't waste your time
To find the skateboard's velocity relative to the sidewalk at this instant, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the jump is equal to the total momentum after the jump.
Before the jump:
The momentum of the boy-skateboard system before the jump can be calculated as the product of their respective masses and velocities. Since the skateboard and the boy have the same velocity (+4.97 m/s) and are moving in the same direction, their momenta add up.
Momentum of the skateboard before the jump = (mass of skateboard) × (velocity of skateboard before the jump)
= (2.39 kg) × (+4.97 m/s)
= 11.8623 kg·m/s
Momentum of the boy before the jump = (mass of the boy) × (velocity of the boy before the jump)
= (39.2 kg) × (+4.97 m/s)
= 195.424 kg·m/s
Total momentum before the jump = Momentum of the skateboard before the jump + Momentum of the boy before the jump
= 11.8623 kg·m/s + 195.424 kg·m/s
= 207.2863 kg·m/s
After the jump:
The boy's velocity after leaving the skateboard (relative to the sidewalk) is given as 6.48 m/s at an angle of 9.54° above the horizontal. We can break down this velocity into horizontal and vertical components using trigonometry.
Horizontal component of the boy's velocity after the jump = (velocity after the jump) × cos(angle)
= 6.48 m/s × cos(9.54°)
= 6.3546 m/s
Vertical component of the boy's velocity after the jump = (velocity after the jump) × sin(angle)
= 6.48 m/s × sin(9.54°)
= 1.1004 m/s
Since there is no friction between the skateboard and the sidewalk, the horizontal component of the boy's velocity after the jump must be equal to the skateboard's velocity relative to the sidewalk at this instant.
Therefore, the skateboard's velocity relative to the sidewalk at this instant is 6.3546 m/s in the same direction as the boy's velocity after the jump.