Question

A $250\text{gm}$ grasshopper moving due south at $20\text{cm/s}$ (in mid air) collides with another $150\text{gm}$ grasshopper moving at a speed of $60\text{cm/s}$ due north. Find the decrease in KE if they move together after collision.

• A. $0.3\text{J}$
• B. $3.0\text{J}$
• C. $0.03\text{J}$
• D. $0.003\text{J}$

Answer 1

The correct option is C $0.03\text{J}$

Mass of grasshopper moving due south
$=m_1=0.25\text{kg}$
& velocity of grasshopper $v_1=-0.20\text{m/s}$
Mass of grasshopper moving due north
$=m_2=0.15\text{kg}$
Velocity of grasshopper $v_2=0.60\text{m/s}$

($+vey-$axis have been considered in due North direction)

Applying $P.C.L.M$ along the $y-$direction:
$P_i=P_f$
$m_1{v}_1+m_2{v}_2=(m_1+m_2)v_f$
$0.25\times (-0.20)+0.15\times \left(0.60\right)=0.40\times v_f$
$\therefore v_f=\frac{0.04}{0.40}=0.1\text{m/s}$
Decrease in K.E after collision:
$\Delta K=K{E}_f-K{E}_i$
$K{E}_f=\frac{1}{2}\times \left(0.40\right)\times (0.1{)}^2=0.002\text{J}$
& $K{E}_i=\frac{1}{2}\times \left(0.25\right)\times (0.2{)}^2+\frac{1}{2}\times (0.15)\times (0.6{)}^2=0.032\text{J}$

$\therefore \Delta KE=0.002-0.032=-0.03\text{J}$
Here, $-ve$ sign represents reduction of kinetic energy after collision.