Question

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposite direction. If the ballis in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is __________
a. 230 N.
b. 400 N.
c. 360 N.
d. 320N,

Answer 1

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball