Question

17. A body of mass 10gm moving with velocity Sms collides with an
body of mass 15 gm moving with 3 msl in the same direction
collision if they stick and move together, the loss of kinetic energy
system is
​

Answer 1

Question :

A body of mass 10gm is moving with a velocity 5 m/s collides with another body of mass 15gm moving with 3 m/s in the same direction. After collision if they stick and move together, the loss of kinetic energy of the system is

Answer :

Given :

• Mass of a body, $\sf m_1$ = 10gm.
• Velocity of a body, $\sf m_2$ = 5 m/s.
• Mass of a body, $\sf m_2$ = 15gm.
• Velocity of second body = 3 m/s.

To find :

• The loss of kinetic energy of the system = ?

Solution :

To find the loss of KE of the system.

First, we need to find the velocity = ?

Using the formula,

$\implies \sf m_1v_1 \ + \ m_2v_2 \ = \ (m_1 \ + \ m_2) v$

Put the given values in the formula.

We get,

$\implies \sf 10 \times 5 \ + \ 15 \times 3 \ = \ (10 + 15) v$

$\implies \sf v \ = \ \dfrac {95}{25}$

$\implies {\red {\bf v \ = \ \dfrac {19}{5} \ m/s}}$

Now,

To find the loss in KE = ?

Using the formula,

$\implies \sf Loss \ in \ KE \ = \ \bigg( \dfrac {1}{2} m_1v_1^2 \ + \ \dfrac {1}{2} m_2v_2^2 \bigg) \ - \ \dfrac {1}{2} mv^2$

Converting the meter in kilometer,

$\implies\sf{\dfrac{1}{1000}\: \bigg\lgroup \dfrac{1}{2} \times 10 \times 5^2 + \dfrac{1}{2} \times 15 \times 3^2 - \dfrac{1}{2} \times 25 \times \dfrac{19}{5} \bigg\rgroup}$

$\implies \sf \dfrac{1}{1000 \times 2} \ \bigg\lgroup 250 \ + \ 135 \ - \ 361 \bigg\rgroup$

$\implies \sf \dfrac {385 \ - \ 361}{2000}$

$\implies \sf \dfrac {29}{2000}$

$\implies \sf \dfrac {12}{1000}$

$\implies {\red {\bf Loss \ in \ KE \ = \ 12 \times 10^{-3} \ J}}$

$\therefore$ The loss of KE of the system is $\sf 12 \times 10^{-3} \ Joules$.