A 7kg object initially at rest explodes Into two parts. One part weigh

1.	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» 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 .

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»

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 .