Question

A 10 g ball falls vertically on a surface at a speed of 5m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 sec.Then what will be the average force exerted by the surface on the ball?

Answer 1

A 10 g ball falls vertically on a surface at a speed of 5m/s andrebounds
with the same speed. The ball remains in contact with the surfacefor 0.01
sec.Then what will be the average force exerted by the surface onthe ball?

Newton's second laws of motion:
F = dp/dt;
F - force;
p - momentum;
t - time;
For average force:
F = Δp/Δt;
Δp = m*v - (-m*v) = 2m*v;
m - mass of the body;
v - velocity;
Δt = 0.01 sec;
F = 2mv/Δt = (2*0.01kg*5m/s)/(0.01sec) = 10 N;
Answer: the average force equals 10 N.