posted by Anonymous on .
A 4.02 kg steel ball strikes a massive wall at 11.8 m/s at an angle of 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle, as shown in the figure below. If the ball is in contact with the wall for 0.174 s, what is the magnitude of the average force exerted by the wall on the ball?
x-axis is along the ball motion and nornal to the wall , y-axis is parallel to the wall
v=v₀ =11.8 m/s ,
linear momentum change is Δp
Δp(y) = m•Δv(y) =m[v-v₀]= 0.
Δp(x)=m•Δv(x) =m[v-(-v₀)] = m•2•v₀ = F•Δt
F= 2•m•v₀/Δt = 2•4.02•10.22/0.174=472.18 N