Question

A block of mass m moving at a velocity v collides with another block of mass 2m . the lighter block comes to rest after collision. find e

Answer 1

for the above collision we will first use momentum conservation

$m* v + 2m* 0 = m* 0 + 2m* v_f$

here

$v_f$ = final speed of block of mass 2m after collision

now from above equation we have

$v = 2*v_f$

now again by the definition of coefficient of restitution

$e = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}$

$e = \frac{\frac{v}{2} - 0}{v - 0}$

$e = \frac{1}{2} = 0.5$

so coefficient of restitution is 0.5

Answer 2

Answer: The value of e is 0.5

Explanation:

Given that,

Mass of first block= m

Mass of second block = 2m

Initial velocity of first block $u_{1} = v$

Initial velocity of second block $u_{2} = 0$

The lighter block comes to rest after collision.

So,

Final velocity of first block $v_{1} = 0$

We know that,

From conservation of momentum

$m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1}+m_{2}v_{2}$

$mv+0=0+2mv_{2}$

$v_{2}= \dfrac{v}{2}$

The coefficient of restitution is

$e = \dfrac{v_{2}-v_{1}}{u_{1}-u_{2}}$

$e = \dfrac{\dfrac{v}{2}-0}{v-0}$

$e = \dfrac{1}{2}$

$e = 0.5$

Hence, the value of e is 0.5