A ball of mass 0.1 kg is dropped from a height of 2 m onto a hard surface. If rebounds to a height of 1.5 m and it is in contact with the surface for 0.05s. Calculate the:
(a) speed with which it strikes the surface.
(b) speed with which it leaves the surface.
(c) change in momentum of the ball.
(d) impules given to the ball on contact with the surface.
(e) average force that the surface exerts on the ball.
To solve this problem, we can use principles of conservation of energy and linear momentum.
(a) To find the speed with which the ball strikes the surface, we can use the principle of conservation of energy. The potential energy at the starting height will convert entirely into kinetic energy at the surface. The potential energy formula is given by:
PE = mgh
Where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
Given:
m = 0.1 kg
g = 9.8 m/s^2
h = 2 m
PE = (0.1 kg) * (9.8 m/s^2) * (2 m) = 1.96 J
Since the potential energy is converted entirely into kinetic energy, we can set the kinetic energy equal to the potential energy:
KE = 1/2 * mv^2
Where v is the velocity or speed of the ball. Solving for v:
1/2 * (0.1 kg) * v^2 = 1.96 J
v^2 = (1.96 J) / (0.05 kg) = 39.2 m^2/s^2
v = √(39.2 m^2/s^2) = 6.26 m/s
Therefore, the speed with which the ball strikes the surface is 6.26 m/s.
(b) To find the speed with which the ball leaves the surface, we can use the principle of conservation of energy again. The ball loses potential energy and gains an equal amount of kinetic energy when rebounding. The maximum potential energy it can gain is at a height of 1.5 m. Using the same formula as in part (a):
PE = (0.1 kg) * (9.8 m/s^2) * (1.5 m) = 1.47 J
Setting this equal to kinetic energy:
1/2 * (0.1 kg) * v^2 = 1.47 J
v^2 = (1.47 J) / (0.05 kg) = 29.4 m^2/s^2
v = √(29.4 m^2/s^2) = 5.42 m/s
Therefore, the speed with which the ball leaves the surface is 5.42 m/s.
(c) The change in momentum of the ball can be calculated using the formula:
Change in momentum = mass * (final velocity - initial velocity)
Given:
m = 0.1 kg
Initial velocity = -6.26 m/s (negative because the ball is dropped from a height)
Final velocity = 5.42 m/s
Change in momentum = (0.1 kg) * (5.42 m/s - (-6.26 m/s))
Change in momentum = (0.1 kg) * (11.68 m/s) = 1.168 kg⋅m/s
Therefore, the change in momentum of the ball is 1.168 kg⋅m/s.
(d) Impulse is defined as the change in momentum of an object over a given time. It can be calculated using the formula:
Impulse = change in momentum = force * time
Given:
Time = 0.05 s
Impulse = (1.168 kg⋅m/s) / (0.05 s) = 23.36 N⋅s
Therefore, the impulse given to the ball on contact with the surface is 23.36 N⋅s.
(e) The average force that the surface exerts on the ball can be calculated using the formula:
Average force = Impulse / time
Given:
Time = 0.05 s
Impulse = 23.36 N⋅s
Average force = (23.36 N⋅s) / (0.05 s) = 467.2 N
Therefore, the average force that the surface exerts on the ball is 467.2 N.