A ball of mass M_ball = 00.6 kg shot with speed v_i = 22 m/s into a barrel of a

Question

A ball of mass M_ball = 00.6 kg shot with speed v_i = 22 m/s into a barrel of a spring gun with mass M_gun = 0.24 kg. The spring gun is initially at rest on a frictionless table. The ball sticks in the barrel at the point of maximum spring compression and in that time the ball and the spring gun have the same leftward velocity. Assume also that there is no friction between the ball and the internal surface of the spring gun. a) What is the velocity of the spring gun and ball system at the time of maximum spring compression? b) What is the elastic potential energy of the spring during that time? What fraction of the initial kinetic energy of the ball has been stored in the spring at that time?

Explanation / Answer

From the given question,

Mball=0.06 kg

speed(vi)=22m/s

Mgun=0.24 kg

a)Using conservation of momentum,

Mball.vi= (Mball+Mgun)vf

0.06*22=(0.06+0.24)vf

vf=4.4 m/s

The velocity of spring gun and ball system at time of maximum compression is 4.4 m/s

b)Using conservation of energy

EPE=elastic potential energy of spring

(1/2) Mballvi2=(1/2)(Mball+Mgun)vf2+ EPE

(1/2)(0.06)*222=(1/2)(0.06+0.24)4.42+EPE

EPE=11.616 J

Elastic potential energy of spring=11.616 J

KE1 of ball=(1/2)(0.06)(222)=14.52 J

Fraction of kinetic energy of ball stored: (14.52-11.66)/14.52=1/5

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