A 43.0-kg boy, riding a 2.20-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, 8.00° 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?
The boy is imparting an impulse with his jump which we know nothing about.
To find the skateboard's velocity relative to the sidewalk at the instant the boy leaves contact with it, we need to consider the law of conservation of momentum.
According to the conservation of momentum, the total momentum before the leap should be equal to the total momentum after the leap.
The momentum of an object is given by its mass multiplied by its velocity. So, we can calculate the initial momentum of the system (boy + skateboard) and equate it to the final momentum.
Initial momentum (before leap):
Total mass of the system = mass of boy + mass of skateboard
= 43.0 kg + 2.20 kg
= 45.20 kg
Total initial momentum = total mass * velocity of the system before the leap
= 45.20 kg * 5.30 m/s
= 239.56 kg·m/s
Final momentum (after leap):
The boy's velocity relative to the sidewalk is given as 6.00 m/s, 8.00° above the horizontal. To find the horizontal component of this velocity, we can use trigonometry:
Horizontal component of velocity = boy's velocity * cos(angle)
= 6.00 m/s * cos(8.00°)
= 5.941 m/s
So, the final momentum of the boy is:
Momentum of the boy = mass of boy * velocity of the boy
= 43.0 kg * 5.941 m/s
= 255.763 kg·m/s
To find the final momentum of the skateboard, we can subtract the momentum of the boy from the total final momentum:
Momentum of the skateboard = Total final momentum - Momentum of the boy
= 239.56 kg·m/s - 255.763 kg·m/s
= -16.203 kg·m/s
Since we are looking for the skateboard's velocity relative to the sidewalk, the negative sign indicates that the skateboard is moving in the opposite direction to the boy's velocity.
Therefore, the skateboard's velocity relative to the sidewalk at this instant is 16.203 m/s in the opposite direction of the boy's velocity.
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 momentum of an object is defined as the product of its mass and its velocity. In this case, the momentum of the boy-skateboard system before the jump is equal to the momentum of the boy alone after the jump.
Given:
Mass of the boy (m1) = 43.0 kg
Mass of the skateboard (m2) = 2.20 kg
Velocity of the boy-skateboard system before the jump (v1) = 5.30 m/s
Velocity of the boy alone after the jump (v2) = 6.00 m/s at an angle of 8.00° above the horizontal
Step 1: Decompose the velocity of the boy after the jump into horizontal and vertical components.
The horizontal component (v2x) of the boy's velocity can be found by multiplying the magnitude of the velocity (v2) by the cosine of the angle (8.00°):
v2x = v2 * cos(8.00°)
Step 2: Apply the conservation of momentum.
The momentum before the jump (p1) is equal to the momentum after the jump (p2):
p1 = p2
Given:
p1 = (m1 + m2) * v1 (Total momentum before the jump)
p2 = m1 * v2x + m2 * v (Momentum after the boy jumps off)
Substituting the given values:
(m1 + m2) * v1 = m1 * v2x + m2 * v
Step 3: Solve for the velocity of the skateboard (v).
Rearrange the equation to solve for v:
v = [(m1 + m2) * v1 - m1 * v2x] / m2
Now let's substitute the given values to calculate the skateboard's velocity relative to the sidewalk:
v = [(43.0 kg + 2.20 kg) * 5.30 m/s - 43.0 kg * v2x] / 2.20 kg
Note: Remember to convert the angle from degrees to radians if your calculator uses radians for trigonometric functions.
Finally, calculate the value of v to find the skateboard's velocity relative to the sidewalk at this instant.