Question

Bob, who has a mass of 78 kg , can throw a 470 g rock with a speed of 33 m/s . The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m.

a) What constant force must Bob exert on the rock to throw it with this speed?

b) If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

Answer 1

Part a.   Use third equation ofmotion

         v^2=   u^2+ 2 * a * s

         v   =   finalspeed   =   30m/s,   u   =   initialspeed   =   0,   a   =   acceleration,   s   =   distancetravelled

         33^2= 0^2+ 2 * a * 1.0

         a   = 1089/ 2   = 544.5 m/s²

Force   F   =   m* a   =   0.470*544.5

F=   256 N

Part b.   By law of conservationof linear momentum

            momentumof rock   =   momentum of Bob

            m* v   =   M * V

So,

         Recoilspeed   V   =   0.470* 33 / 78 =   0.198m/s   =   20 cm/s

I hope help you !!