Question

A body of mass $2\text{kg}$ moving with a velocity $3\text{m/s}$ collides with a body of mass $1\text{kg}$ moving with a velocity of $4\text{m/s}$ in opposite direction. If the collision is head on and completely inelastic, then

• A. Both particles move together with velocity $\left(\frac{2}{3}\right)\text{m/s}$
• B. The momentum of system is $2\text{kg m/s}$ throughout
• C. The momentum of system is $10\text{kg m/s}$
• D. The loss of K.E. of system is $\left(\frac{49}{3}\right)\text{J}$

Answer 1

Answer: (A), (B), (D)

The loss of K.E. of system is $\left(\frac{49}{3}\right)\text{J}$

In completely inelastic collision, bodies coalesce
$m_1{u}_1-m_2{u}_2=(m_1+m_2)V$
$2\times 3-1\times 4=(2+1)V;V=\frac{2}{3}\text{m/s}$
Net momentum of system $P=m_1{u}_1-m_2{u}_2$
$2\times 3-1\times 4=2\text{kg~m/s}$
Loss in KE $=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-\frac{1}{2}(m_1+m_2)V^2$
$=\frac{1}{2}\times 2\times (3{)}^2+\frac{1}{2}\times (1)\times (4{)}^2-\frac{1}{2}\times (2+1)\times {\left(\frac{2}{3}\right)}^2$
$=9+8-\frac{2}{3}=17-\frac{2}{3}=\frac{49}{3}\text{J}$