Question

The man fires an 80 g arrow so that it is moving at 80 m/s when it hits and embeds in a 8.0 kg block resting on ice.

How far will the block slide on the ice before stopping? A 7.1 N friction force opposes its motion.

Answer 1

Step 1: When Arrow hit and embeds in ice block, this is an completely inelastic collision, So

Using Momentum conservation:

Pi = Pf

m1u1 + m2u2 = M*V

m1 = mass of arrow = 80 gm = 0.08 kg

m2 = mass of ice block = 8.0 kg

u1 = initial speed of arrow = 80 m/sec

u2 = initial speed of ice block = 0 m/sec

M = Mass of Arrow + block = 8.08 kg

V = Speed of Arrow+block just after collision = ?

V = (m1u1 + m2u2)/M

V = (0.08*80 + 8.0*0)/8.08

V = 0.792 m/sec

Step 2: Now After collision when block+arrow starts moving, friction force opposes its motion, So de-acceleration of block + arrow will be:

Using newton's 2nd law:

Ff = M*a

a = Ff/M

Ff = -7.1 N (-ve sign since force is in opposite direction of motion)

Which gives

a = -7.1/8.08 = -0.879 m/sec^2

Now Using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*a*S

Vf = final speed of block+arrow = 0 m/sec

Vi = initial speed of block+arrow = 0.792 m/sec

a = -0.879 sec

S = distance traveled by block of ice

S = (Vf^2 - Vi^2)/(2*a)

S = (0^2 - 0.792^2)/(2*(-0.879))

S = 0.357 m = distance traveled by block of ice

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