A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K. E. of colliding body before and after collision will be

Question

A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K. E. of colliding body before and after collision will be (A) 1:1 (B) 2:1 (C) 4:1 (D) 9:1

Tardigrade Admin · Accepted Answer

K. E. of colliding body before collision

= \frac{1}{2} m v^{2}

After collision its velocity becomes

v^{'} = \frac{( m _{1} - m _{2} )}{( m _{1} + m _{2} )} v = \frac{m}{3 m} v = \frac{v}{3}

∴

K. E. after collision

= \frac{1}{2} \cdot \frac{m v ^{2}}{9}

Ratio of kinetic energy

= \frac{K \cdot E _{before}}{K \cdot E _{after}} = \frac{\frac{1}{2} m v ^{2}}{\frac{1 m v ^{2}}{2 \times 9}} = 9 : 1

Q. A body of mass m moving with velocity $v$ collides head on with another body of mass $2 m$ which is initially at rest. The ratio of K. E. of colliding body before and after collision will be

$1 : 1$

$2 : 1$

$4 : 1$

$9 : 1$

Q. A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K. E. of colliding body before and after collision will be

1:1

2:1

4:1

9:1

K. E. of colliding body before collision =21​mv2 After collision its velocity becomes v′=(m1​+m2​)(m1​−m2​)​v=3mm​v=3v​ ∴ K. E. after collision =21​⋅9mv2​ Ratio of kinetic energy =K⋅Eafter ​K⋅Ebefore ​​=2×91mv2​21​mv2​=9:1

Q. A body of mass m moving with velocity $v$ collides head on with another body of mass $2 m$ which is initially at rest. The ratio of K. E. of colliding body before and after collision will be

$1 : 1$

$2 : 1$

$4 : 1$

$9 : 1$