A steel ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.87 m (assume positive direction is upward)
A. Caculate it's velocity just before it strikes the floor
B. calculate it's velocity just after it leaves the floor on its way back up
C. Calculate it's acceleration during contact with the floor if that contact lasts .0800ms
D. How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid
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To calculate the answers to the given questions, we can make use of the laws of motion and conservation of energy.
A. To calculate the velocity of the ball just before it strikes the floor, we can use the principle of conservation of energy. The total mechanical energy (potential energy + kinetic energy) of the ball remains constant throughout its motion. Hence, we equate the potential energy at the initial height to the kinetic energy just before impact.
Potential energy at the initial height:
PE_initial = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the initial height (1.95 m).
Kinetic energy just before impact:
KE_final = 0.5 * m * v²
where v is the velocity just before the ball strikes the floor.
Setting PE_initial equal to KE_final, we get:
m * g * h = 0.5 * m * v²
Simplifying the equation and solving for v:
v = sqrt(2 * g * h)
Substituting the given values:
v = sqrt(2 * 9.8 * 1.95) ≈ 6.29 m/s
Therefore, the velocity of the ball just before it strikes the floor is approximately 6.29 m/s.
B. To calculate the velocity of the ball just after it leaves the floor on its way back up, we can again use the principle of conservation of energy. The total mechanical energy just after leaving the floor is equal to the kinetic energy at the maximum height.
Kinetic energy at the maximum height:
KE_final = 0.5 * m * v²
where v is the velocity just after leaving the floor.
Setting KE_final equal to the potential energy at the maximum height:
0.5 * m * v² = m * g * h
Simplifying and solving for v:
v = sqrt(2 * g * h)
Substituting the given values:
v = sqrt(2 * 9.8 * 1.87) ≈ 6.07 m/s
Therefore, the velocity of the ball just after it leaves the floor on its way back up is approximately 6.07 m/s.
C. To calculate the acceleration during contact with the floor, we need to use the formula for acceleration:
Acceleration = (change in velocity) / (time taken)
Given that the contact lasts 0.0800 ms (0.0800 seconds) and the change in velocity is the difference between the velocity just before and just after the collision, we can calculate the acceleration using the formula above.
Acceleration = (v_after - v_before) / time
Substituting the given values:
Acceleration = (6.07 m/s - (-6.29 m/s)) / 0.0800 s ≈ 156.25 m/s²
Therefore, the acceleration during contact with the floor is approximately 156.25 m/s².
D. To determine how much the ball compresses during its collision with the floor, we need additional information such as the coefficient of restitution or the elasticity of the ball and the floor. Without this information, we cannot determine the amount of compression.