A 42.4-kg boy, riding a 2.29-kg skateboard at a velocity of +5.37 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 6.30 m/s, 9.09° 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 will be in m/s.
To find the skateboard's velocity relative to the sidewalk at the instant the boy jumps, we can use the law of conservation of momentum. According to this law, the total momentum before the boy jumps must be equal to the total momentum after the jump.
Let's start by calculating the total momentum before the jump. The momentum of an object is defined as the product of its mass and velocity. So, the momentum of the boy-skateboard system before the jump is:
Momentum before = (mass of the boy × velocity of the boy) + (mass of the skateboard × velocity of skateboard)
= (42.4 kg × +5.37 m/s) + (2.29 kg × 0 m/s) [since the skateboard is not moving horizontally]
Now, let's calculate the momentum after the jump. The momentum after the jump is solely due to the boy since the skateboard is not in contact with the boy anymore. The momentum after the jump is given as:
Momentum after = mass of the boy × velocity of the boy after the jump
= 42.4 kg × 6.30 m/s [we are given this velocity in the question]
Since the law of conservation of momentum states that the momentum before the jump is equal to the momentum after the jump, we can set these two equations equal to each other and solve for the velocity of the skateboard after the jump.
(42.4 kg × +5.37 m/s) + (2.29 kg × 0 m/s) = 42.4 kg × 6.30 m/s
Simplifying the equation further:
227.488 + 0 = 266.52
Rearranging the equation:
266.52 - 227.488 = 0
Now, we can solve for the difference in velocity:
Delta velocity = (266.52 - 227.488) / (42.4 kg)
Simplifying the equation:
Delta velocity = 39.032 / 42.4
Delta velocity ≈ 0.921 m/s
Finally, we need to determine the direction of the skateboard's velocity relative to the sidewalk at this instant. We are given that the boy's velocity after the jump is 9.09° above the horizontal. Since the skateboard and the boy are initially moving in the same direction, we can conclude that the skateboard's velocity relative to the sidewalk is also 9.09° above the horizontal and in the same direction as the boy's velocity. Therefore, the skateboard's velocity relative to the sidewalk at this instant is approximately 0.921 m/s, 9.09° above the horizontal in the direction of the boy's velocity.