A boy places a fire cracker of negligible mass in an empty can of 40 g mass. He plugs the end with a wooden block of mass 200 g. After igniting the fire cracker, he throws the can straight up. It explodes at the top of its path. If the block shoots out with a speed of 3 m s⁻¹, how fast will the can be going? 

(Ans: 15 m s⁻¹)



Given:

Mass of can = m₁ = 40 g  = 0.04 kg

Mass of wooden block = m₂ = 200 g  = 0.2 kg

Velocity of wooden block =  V₂ = 3 m s⁻¹   


To Find:

Velocity of can =  V₁ = ?   


Solution:


By using Law of conservation of momentum

Momentum before explosion = Momentum after explosion  ------------ (1)

Momentum before explosion = 0 Kg  m s⁻¹ 
( Because at the highest top the system became at rest)

Momentum after explosion = m₁v₁ + m₂v₂

so equation (1)

0 = m₁v₁ + m₂v₂

or

v₁ = -`\frac {m₂v₂}{m₁}`

by putting the corresponding values

v₁ = -`\frac {0.2 kg ｘ 3 m s⁻¹ }{0.04 kg}`

v₁ = -15 m s⁻¹ -------------Ans.

the negative shows the direction of the can that is in opposite to the direction of wooden block.


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