Question

A 7kg object initially at rest explodes Into two parts. One part weighing 2kg goes east at the velocity of 20m/s,find the velocity of the second part

Answer 1

Given :

• The mass of the object : 7 kg
• It was stationary initially .
• One part weight = 20 kg
• Velocity of part one = 20 m/s in East

To find :

• Velocity of second part

Solution:

Mass of the second part = Mass of object - Mass of first part

= 7 - 2

= 5 kg

Velocity of the body initially = 0 .

The formula of momentum is given by :

$\boxed{\tt momentum = mass \times velocity}$

p = mv

p = 7 × 0

p = 0 kgm/s

So , the initial momentum of the body is 0 .

Momentum of the first part is :

p = mv

p = 2 × 20

p = 40 kgm/s

According to law of conservation of momentum ,

Initial momentum = Final momentum

Therefore , the momentum of the body will be equal to the sum of the momentums of the two parts .

0 = 40 + momentum of second

Momentum of second part = -40 kgm/s

This -ve sign shows that the direction of momentum will be opposite to the direction of momentum of first part ,i.e. towards west so that the momentum remains conserved .

Therefore , the magnitude of the momentum of second part = 40 kgm/s .

• Mass of second part = 5 kg
• Momentum of second part = 40kgm/s
• Velocity of second part = ?

Using formula of momentum :

$\boxed{\tt velocity = \dfrac{momentum}{mass}}$

v = 40/5

v = 8m/s

So , the second part will move with the velocity of 8 m/s in the west direction .