*62. 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, 9.50° 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.
To find the skateboard's velocity relative to the sidewalk at the instant the boy leaves contact with it, we can use the principle of conservation of momentum.
The total momentum before the boy jumps can be calculated by adding the momenta of the boy and the skateboard. Since the direction is important in this problem, we will consider motion to the right as positive.
The momentum of the boy (m_boy * v_boy) can be calculated by multiplying the mass of the boy (m_boy = 40.0 kg) by his velocity relative to the sidewalk (v_boy = 6.00 m/s at an angle of 9.50° above the horizontal). Since we need the horizontal component of this velocity, we can calculate it using trigonometry.
The horizontal velocity component (v_boy_horizontal) can be found by multiplying the magnitude of the velocity (6.00 m/s) by the cosine of the angle (9.50°). So, v_boy_horizontal = 6.00 m/s * cos(9.50°).
The momentum of the skateboard (m_skateboard * v_skateboard) can be calculated by multiplying the mass of the skateboard (m_skateboard = 2.50 kg) by its velocity relative to the sidewalk.
Since the total momentum before the boy jumps is conserved, it will be equal to the total momentum after the jump, where the skateboard will have a new velocity.
The momentum after the boy jumps can be calculated by multiplying the total mass (m_boy + m_skateboard) by the final velocity of both the boy and the skateboard. We are given the final velocity of the boy relative to the sidewalk (6.00 m/s), which means the new velocity of the skateboard relative to the sidewalk is also 6.00 m/s.
Now we can set up the equation using the principle of conservation of momentum:
(m_boy * v_boy_horizontal) + (m_skateboard * v_skateboard) = (m_boy + m_skateboard) * (final velocity)
Plug in the known values:
(40.0 kg) * (6.00 m/s * cos(9.50°)) + (2.50 kg) * v_skateboard = (40.0 kg + 2.50 kg) * (6.00 m/s)
Now we can solve for v_skateboard:
(240.0 kg m/s * cos(9.50°)) + (2.50 kg) * v_skateboard = (42.5 kg) * (6.00 m/s)
Rearranging the equation to solve for v_skateboard:
(2.50 kg) * v_skateboard = (42.5 kg) * (6.00 m/s) - (240.0 kg m/s * cos(9.50°))
v_skateboard = [(42.5 kg) * (6.00 m/s) - (240.0 kg m/s * cos(9.50°))] / (2.50 kg)
Now, substitute the given values and calculate:
v_skateboard = [(42.5 kg) * (6.00 m/s) - (240.0 kg m/s * cos(9.50°))] / (2.50 kg)
v_skateboard = [255.0 kg m/s - (240.0 kg m/s * cos(9.50°))] / (2.50 kg)
v_skateboard = [255.0 kg m/s - (240.0 kg m/s * cos(9.50°))] / (2.50 kg)
v_skateboard ≈ -0.628 m/s
Therefore, the skateboard's velocity relative to the sidewalk at the instant the boy leaves contact with it is approximately -0.628 m/s.
To find the skateboard's velocity relative to the sidewalk at this instant, we can use the law of conservation of momentum. According to this law, the total momentum before and after an interaction remains constant, assuming no external forces are acting on the system.
Let's denote the velocity of the skateboard before the boy jumps as v1 and the velocity of the skateboard after the boy jumps as v2.
Given:
- Mass of the boy (m1) = 40.0 kg
- Mass of the skateboard (m2) = 2.50 kg
- Velocity of the skateboard before the boy jumps (v1) = +5.30 m/s
- Velocity of the boy relative to the sidewalk after the jump (v3) = 6.00 m/s
We want to find the velocity of the skateboard relative to the sidewalk after the jump (v2).
To solve for v2, we can use the conservation of momentum equation:
m1v1 + m2v1 = m1v3 + m2v2
Substituting the given values:
(40.0 kg)(5.30 m/s) + (2.50 kg)(5.30 m/s) = (40.0 kg)(6.00 m/s) + (2.50 kg)(v2)
Simplifying the equation:
212 kg·m/s + 13.25 kg·m/s = 240 kg·m/s + 2.50 kg·v2
225.25 kg·m/s = 240 kg·m/s + 2.50 kg·v2
Rearranging the equation to solve for v2:
2.50 kg·v2 = 225.25 kg·m/s - 240 kg·m/s
2.50 kg·v2 = -14.75 kg·m/s
Dividing both sides by 2.50 kg:
v2 = -14.75 kg·m/s ÷ 2.50 kg
v2 = -5.90 m/s
Therefore, the skateboard's velocity relative to the sidewalk at this instant is -5.90 m/s. The negative sign indicates that the skateboard is moving in the opposite direction.