A girl of mass 55 kg throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with the same speed. The ball is in contact with the wall 0.05 s. What is the average force exerted on the wall by the ball?
A) 400 N
B) 800 N
C) 55,000 N
D) 27,500 N
E) 13,750 N
800N
(Avg force) x (contact time) = momentum change
= 55*50 kg*m/s
Avg. force = 0.8*50/.05 = ____ N
One of the choices works.
To find the average force exerted on the wall by the ball, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the object's mass multiplied by its acceleration.
First, let's find the acceleration of the ball. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
In this case, the initial velocity (u) of the ball is 25 m/s, and the final velocity (v) is also 25 m/s since the ball bounces back with the same speed. The time taken (t) is given as 0.05 s. Therefore, we can rearrange the equation to solve for the acceleration (a):
v = u + at
25 = 25 + a * 0.05
25 - 25 = a * 0.05
0 = a * 0.05
a = 0 / 0.05
a = 0 m/s^2
Since the acceleration is 0 m/s^2, it means that the ball does not experience any acceleration during the contact with the wall.
Now, let's calculate the force exerted on the wall. We can use the equation F = m * a, where F is the force, m is the mass of the ball, and a is the acceleration.
The mass of the ball is given as 0.8 kg. Since the acceleration is 0 m/s^2, the force exerted on the wall is:
F = m * a
F = 0.8 kg * 0 m/s^2
F = 0 N
Therefore, the average force exerted on the wall by the ball is 0 N.
The correct answer is None of the above.