Question

A bomb of mass
3m kg explodes into
two pieces of mass
m kg and 2m kg. If
the velocity of m kg
mass is 16 m/s, then
the total kinetic
energy released in
the explosion is​

Answer 1

Given :

A bomb of mass  3m kg explodes into  two pieces of mass  m kg and 2m kg. If  the velocity of m kg  mass is 16 m/s

To Find :

Total Kinetic energy released in the explosion

Solution :

Here , a mass of 3m kg which is in rest explodes into two pieces of mass m kg and 2m kg . Velocity of m kg is 16 m/s .

Apply conservation of momentum ,

⇒ (3m)(0) = m(16) + (2m)v

⇒ 2mv = - 16m

⇒ v = - 8 m/s  $\pink{\bigstar}$

Note : - ve sign denotes opposite direction of 2m's velocity with m's velocity

Total kinetic energy released in the explosion ,

$\to \sf K.E =\dfrac{1}{2}m(16)^2+\dfrac{1}{2}(2m)(-8)^2\\\\\to \sf K.E=128m+64m\\\\\to \sf \pink{K.E=192m\ J}\ \; \bigstar$

Answer 2

Given : A bomb of mass 3 m kg explodes into two pieces of mass (m) kg and 2m kg. If the velocity of m kg mass is 16 m/s.

　

To find : What is the total kinetic energy released in

the explosion.

　

Applying formula:

→ KE = (m¹ v¹ + m² v²)

Calculations :

→ v = 2/-16

→ v = -8 m/s

　

→ KE = 1/2 (16)² + 1/2 (2) (-8)²

→ KE = 128 + 64

→ KE = 192 mj

　

Therefore, 192 mj is the total kinetic energy released in the explosion.