Question

Bob, who has a mass of 80kg , can throw a 480g rock with a speed of 26 m/s . The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0m .

Part A : What constant force must Bob exert on the rock to throw it with this speed?

Part B : If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

Answer 1

Part A.

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

Given that U = Initial speed of hand = 0 m/sec

V = final speed of hand = 26 m/sec

d = distance traveled by hand = 1.0 m

So,

a = acceleration of hand = (V^2 - U^2)/(2*d)

a = (26^2 - 0^2)/(2*1.0)

a = 338 m/sec^2

Now Using newton's 2nd law:

Force exerted on the rock will be:

F_net = m*a

m = mass of rock = 480 gm = 0.480 kg

So

F_net = 0.480*338

F_net = 162.24 N

Part B.

Using momentum conservation:

Pi = Pf

m1u1 + m2u2 = m1v1 + m2v2

m1 = mass of Bob = 80 kg

u1 = initial speed of bob = 0 m/sec

m2 = mass of rock = 480 gm = 0.480 kg

u2 = initial speed of rock = 0 m/sec

So Pi = 80*0 + 0.480*0 = 0 kg-m/sec

which means Pf = 0

v1 = final speed of bob = ? m/sec

v2 = final speed of rock = 26 m/sec

So,

80*v1 + 0.480*v2 = 0

v1 = -0.480*26/80

v1 = -0.156 m/sec = recoil speed of Bob after releasing the rock (here -ve sign shows that this is recoil speed)

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