Question

a sand bag of mass M is suspended by rope. A
bullet of mass m is fired at it with speed v and the
bullet gets embedded in it. The loss of kinetic
energy in the collision is
Mmy?
(1) 2(M + m)
Mv2
(2) 2(M + m)
m²v²
(3) 2(M + m)
(4) 3 (M + m)Y

please provide explanation
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Answer 1

Explanation:

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Please refer to attachment .

Approach

1. We can see that we will have to find final velocity by applying laws of conservation of momentum.

2. Since initially , the momentum of bullet was maximum & bag had zero momentum so we will subtract the kinetic energy of (bullet-bag) system from kinetic energy of bullet .

Note : I have substituted the value of v' (final velocity) while calculating lost kinetic energy.

Answer 2

Answer:

Mmv^2/2(m+M).

Explanation:

We know that for change in energy we have the initial momentum = final momentum which will be (M+m)u=mv.

The ratio of the final kinetic energy to the initial kinetic energy is given by the 1/2(m+M)u^2/1/2mv^2/ which will be (M+m)/m(u/v)^2 which will be (M+m)/m*/(m/M+m)^2 or m/(m+M).

So, finally the kinetic energy loss will be initial kinetic energy - the final kinetic energy or initial kinetic energy(1- final kinetic energy/initial kinetic energy) which will be (1/2mv^2)*M/m+M or Mmv^2/2(m+M).