Question

A vehicle weight 2 kN skids through a distance before colliding with another parked vehicle of weight 2 kN. After collision, both the vehicle skid through a distance of 10 m before stopping. If the coefficient of friction is 0.5 then speed of the vehicles after the collision is

• A. 4.63 m/sec
• B. 3.16 m/sec
• C. 3.54 m/sec
• D. 9.9 m/sec

Answer 1

The correct option is D 9.9 m/sec

After collision both the vehicles move together

Let 'V' be the speed of the vehicles after collision.

$Kineticenergy=\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2\times 9.81}(2+2)V^2=0.204V^2$

When the vehicles stop, kinetic energy = 0
Loss in kinetic energy
$=0.204V^2$
= workdone against frictional force
$0.204V^2=(w_1+w_2)fS$
$0.204V^2=(2+2)\times 0.5\times 10$
$V=9.9m/sec$