A 38.0-kg boy, riding a 2.37-kg skateboard at a velocity of +5.07 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 6.06 m/s, 10.35° 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.
x momentum after = x momentum before
after
p = 38 (6.06 cos 10.5) + 2.37 v
before
p = (38+2.37)(5.07)
set equal, solve for v which will probably be negative (kid goes forward, board goes backward)
Damon, Thank you so much
To solve this problem, we can use the principle of conservation of momentum. The total momentum of the system (boy + skateboard) before and after the jump must be the same. The momentum is given by the product of mass and velocity:
Initial momentum = Final momentum
The initial momentum consists of the boy's momentum and the skateboard's momentum before the jump.
The boy's momentum before the jump is calculated as:
Momentum of boy before jump = mass of boy * velocity of boy before jump
Momentum of boy before jump = 38.0 kg * 5.07 m/s (considering the velocity as positive)
Momentum of boy before jump = 192.66 kg·m/s
The skateboard's momentum before the jump is calculated as:
Momentum of skateboard before jump = mass of skateboard * velocity of skateboard before jump
Momentum of skateboard before jump = 2.37 kg * 5.07 m/s (considering the velocity as positive)
Momentum of skateboard before jump = 11.9859 kg·m/s
(we will use the full value temporarily and round it later)
Therefore, the total initial momentum is:
Initial momentum = Momentum of boy before jump + Momentum of skateboard before jump
Initial momentum = 192.66 kg·m/s + 11.9859 kg·m/s (considering the velocities as positive)
Initial momentum = 204.6459 kg·m/s
Now, let's calculate the final momentum of the system after the jump. The final momentum consists of the boy's momentum and the skateboard's momentum after the jump.
The boy's momentum after the jump is calculated as:
Momentum of boy after jump = mass of boy * velocity of boy after jump
Momentum of boy after jump = 38.0 kg * 6.06 m/s (considering the velocity as positive)
Momentum of boy after jump = 230.28 kg·m/s (rounded to two decimal places)
The direction of the boy's velocity is given as 10.35° above the horizontal. Since the skateboard's velocity is in the same direction, we can use simple vector addition to determine the skateboard's velocity relative to the sidewalk.
The x-component (horizontal component) of the final velocity is given by:
vx = v * cos(θ)
vx = 6.06 m/s * cos(10.35°)
vx ≈ 5.8683591 m/s (rounded to six decimal places)
The y-component (vertical component) of the final velocity is given by:
vy = v * sin(θ)
vy = 6.06 m/s * sin(10.35°)
vy ≈ 1.0780437 m/s (rounded to six decimal places)
Therefore, the skateboard's velocity relative to the sidewalk after the jump is:
Velocity of skateboard in x-component = vx ≈ 5.8684 m/s
Velocity of skateboard in y-component = vy ≈ 1.0780 m/s
Thus, the skateboard's velocity relative to the sidewalk at this instant is approximately 5.8684 m/s in the horizontal direction and 1.0780 m/s in the vertical direction, taking into account the correct algebraic sign for each component.
To find the skateboard's velocity relative to the sidewalk at the instant the boy jumps off, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.
In this case, the boy and the skateboard form an isolated system since no external forces are mentioned. Therefore, the total momentum before the boy jumps off will be equal to the total momentum after he jumps off.
The total momentum before the boy jumps off can be calculated by multiplying the mass of the boy and the skateboard by their initial velocity.
Total momentum before = (mass of the boy + mass of the skateboard) × initial velocity
= (38.0 kg + 2.37 kg) × 5.07 m/s
= 200.7 kg·m/s
Now, let's consider the velocity of the boy after he jumps off. The velocity of the boy can be separated into horizontal and vertical components. The horizontal component can be found using the equation:
horizontal component = magnitude of velocity × cos(angle above horizontal)
horizontal component = 6.06 m/s × cos(10.35°)
horizontal component = 6.06 m/s × 0.9813
horizontal component = 5.9462 m/s
Since there are no horizontal external forces acting on the system, the horizontal component of the boy's momentum after he jumps off must be equal to the horizontal component of the total momentum before jumping off. Therefore, we can write:
horizontal momentum before = (mass of the boy + mass of the skateboard) × initial velocity
horizontal momentum before = (38.0 kg + 2.37 kg) × 5.07 m/s
horizontal momentum before = 200.7 kg·m/s
horizontal momentum after = (mass of the boy) × (horizontal component of velocity after jumping off)
horizontal momentum after = 38.0 kg × 5.9462 m/s
horizontal momentum after = 226.0156 kg·m/s
Now, we can calculate the horizontal momentum of the skateboard after the boy jumps off by subtracting the boy's horizontal momentum after jumping off from the total horizontal momentum after jumping off.
Horizontal momentum of the skateboard after = horizontal momentum after - horizontal momentum of the boy after
Horizontal momentum of the skateboard after = 226.0156 kg·m/s - 226.0156 kg·m/s
Horizontal momentum of the skateboard after = 0 kg·m/s
Since the horizontal momentum of the skateboard after the boy jumps off is zero, the skateboard's velocity relative to the sidewalk at this instant is 0 m/s in the horizontal direction. The algebraic sign is positive because it is in the same direction as the boy's initial velocity.