Question

An object of mass 2kg falls from the height of 5m and rebounds to a vertical height of 2m. Calculate the Change in momentum. If the object is in contact with the floor for 0.2 seconds, calculate the force experienced by the object from the floor.

Answer 1

The change in momentum is 20 m/s downward. The force experienced by the object from the floor is 100 N upward.

Step-by-step explanation:

The change in momentum (Δp) is calculated using the formula Δp = m * Δv, where m is the mass of the object and Δv is the change in velocity. The initial velocity (u) is 0, and the final velocity (v) is calculated using the kinematic equation
`v^2` = u^2 + 2as, where a is the acceleration due to gravity and s is the displacement. Substituting the values, we find v = √(2 * 9.8 * 5) = 10 m/s. Therefore, Δv = v - u = 10 m/s.

Now, calculating Δp, Δp = 2 kg * 10 m/s = 20 kg·m/s. The negative sign indicates that the change in momentum is downward.

To find the force experienced by the object from the floor, we use Newton's second law, F = Δp / Δt, where Δt is the time of contact. Substituting the values, F = 20 kg·m/s / 0.2 s = 100 N. The upward force of 100 N is exerted by the floor on the object during the 0.2 seconds of contact.

In summary, the change in momentum is 20 m/s downward, representing the impulse experienced by the object. The force exerted by the floor is 100 N upward, equal and opposite to the change in momentum, as per Newton's third law.