Question

A body of mass 300 kg kept at rest breaks into two parts due to internal forces. One part of 200kg is found to move at a speed of 12m/s towards east. What will be the velocity of the other part?​

Answer 1

Answer:

24 m s^-1 towards east.

Explanation:

Mass of one part m

1

=200 gm

Its velocity v

1

=12 m/s towards east

So, mass of other part m

2

=300−200=100 gm

Let its velocity be v

2

.

Applying conservation of linear momentum : P

f

−P

i

Or m

1

v

1

+m

2

v

2

=0

Or 200×12+100×v

1

=0

⟹ v

2

=−24 m/s

So, the second part will fly off towards west with a velocity of 24 m/s.

Answer 2

Answer:

To Find:-

Velocity of Second part

Solution:-

Let the first Part be M1

second part = M2

mass of M1=200kg

Given,the object is at rest with a mass of 300 kg

200 kg,that is M1 moved towards east

therefore the remaining is

300-200=100

therefore

Mass of M1=100 kg

Let the velocity of M1= V1

Velocity of M2= V2

We know

momentum=mass×velocity

Also we know

According to Conservation Of Linear Momentum

Momentum Before a system=Momentum after the system

Applying it here

Momentum Before the System= 300 kg × 0m/s (at rest)

Momentum Before the System= 0 kgm/s

Momentum After the System

M1×V1+M2×V2

200×12+100×V2

Momentum Before a system=Momentum after the system

0=200×12+100×V2

2400+100V2=0

100V2=0-2400=-2400

V2=-2400÷100

V2=-24

Therefore the velocity of of the second object is -24m/s

The negative sign implies that, it's moving opposite to the Direction of M1

that is

Second object will move with a speed of 24m/s to west