A 500 g rubber ball is thrown to the right at a wall at a speed of 5.00 m / s. The ball bounces to the left at a speed of 4.50 m / s. If the collision between the ball and the wall lasted 0.250 s, what is the force exerted by the wall on the ball?
F=ΔpΔt=p2−p1Δt=mv2−(−mv1)Δt=m(v2+v1)Δt=0.5⋅9.50.25=19 N.F=\frac{\Delta p}{\Delta t}=\frac{p_2-p_1}{\Delta t}=\frac{mv_2-(-mv_1)}{\Delta t}=\frac{m(v_2+v_1)}{\Delta t}=\frac{0.5\cdot 9.5}{0.25}=19~N.F=ΔtΔp=Δtp2−p1=Δtmv2−(−mv1)=Δtm(v2+v1)=0.250.5⋅9.5=19 N.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments