Question

A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it .The kinetic energy of the composite block will be
a-1/2mv^2(m/m+M)
b-1/2mv^2(M/m+M)
c-1/2mv^2(2m/m+M)
d-1/2mv^2(m/M)​

Answer 1

Consider, the velocity of bullet just before hitting the block is v m/s.

Now, The bigger block of mass M is in rest,

hence u = 0 m/s.

Since, No external Force acts on the system, We can say that the momentum of the system remains conserved throughout the collision.

Applying Conservation of Momentum,

Initial Momentum = Final Momentum

$→ \: P_{i} = P_{f}$

Mass of bullet = m

Mass of bigger block = M

initial velocity of bullet = v m/s

Initial velocity of M = 0

Since, The bullet gets embedded into the block 'M',

the total mass of system after collision becomes = (m + M) ......(i)

Let the combined Velocity of the system

(m + M) be = Vc ......(ii)

Now,

$→ \: mv + M(0) = (m + M)V_{c}$

$→ \: V_{c} = \frac{mv}{(m + M)}$

Therefore, Kinetic Energy of the Composite system becomes :

$→ \: \frac{1}{2}M_{system} {v_{c}}^{2}$

$→ \: \frac{1}{2}(m + M) {(\frac{mv}{(m + M)})}^{2} \\ \\ → \: \frac{1}{2} \frac{{m}^{2} {v}^{2} }{(m + M)}$

$→ \: \frac{1}{2}m {v}^{2}( \frac{m}{(m + M)} )$

Which is Option (a)

Hope this helps!