Asked by sugin
A 200g tennis ball approaches a racket at 20m/s, is in contact with the racket for 0.008s, and the ball rebounds at 4m/s. Find the average force that the racket exerted on the ball.
Answers
Answered by
Elena
Change of the linear momentum of the ball is
Δp= mv1-(-mv2) =m(v1+v2) =0.2(20+4) =4.8 kg•m•s^-1
Newton’s 2 Law
F=Δp/Δt =4.8/0.008=600 N
Δp= mv1-(-mv2) =m(v1+v2) =0.2(20+4) =4.8 kg•m•s^-1
Newton’s 2 Law
F=Δp/Δt =4.8/0.008=600 N
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