Question #139977

During an impact time lasting approximately 5.0*10^-4 s a golf club exerts an average force of about 500N on the golf ball. What linear impulse is delivered to the ball as a result?

Expert's answer

We can find the linear impulse delivered to the ball from the impulse-momentum change equation:


FavgΔt=mΔv,F_{avg}\Delta t=m\Delta v,FavgΔt=mvfmvi,F_{avg}\Delta t=mv_f-mv_i,FavgΔt=mvf=p,F_{avg}\Delta t=mv_f=p,

here, mm is the mass of the golf ball, vi=0v_i=0 is the initial velocity of the golf ball, vfv_f is the final velocity of the golf ball, Favg=500 NF_{avg}=500\ N is the average force that the golf club exerts on the golf ball, Δt=5.0104 s\Delta t=5.0\cdot10^{-4}\ s is the impact time during which the golf club interacts with the golf ball, pp is the linear impulse delivered to the ball.

Then, from this equation we can find the linear impulse delivered to the ball:


p=FavgΔt=500 N5.0104 s=0.25 kgms.p=F_{avg}\Delta t=500\ N\cdot 5.0\cdot10^{-4}\ s=0.25\ \dfrac{kgm}{s}.

Answer:

p=0.25 kgms.p=0.25\ \dfrac{kgm}{s}.


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