Question

A bag of mass M hangs by a long thread and a
bullet (mass m) comes horizontally with velocity v
and get caught in the bag. Then for the combined
system (bag + bullet)
mMv
(1) Momentum is
(M + m)
(2) Kinetic energy is - Mv2
(3) Momentum is
mv(M + m)
M
m²v2
(4) Kinetic energy is. m^2v^2/2(M+m)​

Answer 1

Answer:

Ans 4

$ke = \frac{1}{2} \frac{ {m}^{2} {v}^{2} }{total \: mass}$

Explanation:

From conservation of momentum

mV = (M + m) v(final)

mV/(M+m) = v final

$ke = \frac{1}{2} mass \: \times {v}^{2} \\ ke = \frac{1}{2} total \: mass \times ( \frac{ {m}^{2} {v}^{2} }{ {total \: mass}^{2} }) \\ ke = \frac{1}{2} \frac{ {m}^{2} {v}^{2} }{(m + m)}$

Answer 2

Explanation:

hope it helps u...........