Question

1. Two bodies with kinetic energies in ratio of 4 : 1 are moving with equal linear momentum .the ratio pf their massses is ?

2. A shell of mass m moving with velocity v suddenly breaks into 2 pieces .the part having mass m/4 remains stationary . the velocity of other part will be ?

Answer 1

1. We know that,
Kinetic Energy= 1/2mv²
K.E= mv/2×v--------(1)

Now
p(momentum)= m×v
v=p/m-------------(2)

Putting (2) in (1)
mv/2×p/m= K.E
p/2×p/m= K.E
p²/2m= K.E

Let the mass of one object be m and momentum p.
Let the mass of another object be m' and momentum p'.

Kinetic Energy of Object 1= p²/2m

Kinetic Energy of Object 2= p'²/2m'

Kinetic Energy of Object 1/Kinetic Energy of
Object 2= 4/1

Kinetic Energy of Object 1/Kinetic Energy of
Object 2= p²/2m÷p'²/2m'

As p= p’, p²=p'²
4/1=2m/2m’
4/1=m/m'
m:m'= 4:1

2. Here, we will use the formula of m1u1+m2u2= m1v1+m2v2
Here, m1u1+m2u2= (m1+m2)(v)= mv
mv= m1v1+m2v2
As Object with mass m1 is stationary after explosion,
mv= m2v2
m2= m-m1= m-m/4= 3m/4
mv= (3m/4)(v2)
(mv)(4)/3m= v2
4mv/3m=v2
4v/3 m/s = v2

Hope it helps! Please mark it as 'Brainliest'!

Answer 2

1. We know that,

Kinetic Energy= 1/2mv²

K.E= mv/2×v--------(1)

Now

p(momentum)= m×v

v=p/m-------------(2)

Putting (2) in (1)

mv/2×p/m= K.E

p/2×p/m= K.E

p²/2m= K.E

Let the mass of one object be m and momentum p.

Let the mass of another object be m' and momentum p'.

Kinetic Energy of Object 1= p²/2m

Kinetic Energy of Object 2= p'²/2m'

Kinetic Energy of Object 1/Kinetic Energy of

Object 2= 4/1

Kinetic Energy of Object 1/Kinetic Energy of

Object 2= p²/2m÷p'²/2m'

As p= p’, p²=p'²

4/1=2m/2m’

4/1=m/m'

m:m'= 4:1

2. Here, we will use the formula of m1u1+m2u2= m1v1+m2v2

Here, m1u1+m2u2= (m1+m2)(v)= mv

mv= m1v1+m2v2

As Object with mass m1 is stationary after explosion,

mv= m2v2

m2= m-m1= m-m/4= 3m/4

mv= (3m/4)(v2)

(mv)(4)/3m= v2

4mv/3m=v2

4v/3 m/s = v2

Hope it helps! Please mark it as 'Brainliest'!