A 200g ball thrown at 2 m/s bounces off a wall and returns to the thrower at 1 m/s. If the ball is in contact with the wall for .1 seconds, what is the average force exerted by the wall on the ball?600 N6 N3 N1 N
Question
A 200g ball thrown at 2 m/s bounces off a wall and returns to the thrower at 1 m/s. If the ball is in contact with the wall for .1 seconds, what is the average force exerted by the wall on the ball?600 N6 N3 N1 N
Solution
To solve this problem, we need to use the formula for force, which is F = m*a, where F is force, m is mass, and a is acceleration.
First, we need to find the change in velocity. The ball was initially moving at 2 m/s and then it returned at 1 m/s in the opposite direction. So, the change in velocity is final velocity - initial velocity = -1 m/s - 2 m/s = -3 m/s.
Next, we need to find the acceleration. Acceleration is the change in velocity divided by the time it takes for this change to occur. So, acceleration = change in velocity / time = -3 m/s / 0.1 s = -30 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial motion.
Finally, we can find the force. The mass of the ball is 200 g, but we need to convert this to kg because the standard unit of mass in physics is kilograms. So, mass = 200 g = 0.2 kg.
Now we can calculate the force: F = m*a = 0.2 kg * -30 m/s^2 = -6 N. The negative sign indicates that the force is in the opposite direction of the initial motion, which makes sense because the wall is pushing the ball back towards the thrower.
So, the average force exerted by the wall on the ball is 6 N.
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