Question

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

Answer 1

Answer:

Explanation: In this problem we will have a mass m strike another mass of 4m in a head on collision. This will be found out as elastic or inelastic collision depending upon the e or coefficient of restitution value. If it is 1 it's elastic or otherwise inelastic if ranges from 0 to 1.

Now as we know that momentum is conserved and the masses of the object will remain same before and after collision.

We have the formula as

MaVia + MbVib = MaVfa + MbVfb

Ma and Mb are masses given as m and 4m

Via and Vib are inital velocities of a and by which is v and 0 respectively (object at rest)

Vfa and Vfb are 0(comes to rest) and to be found out respectively

We will have in simplified way ,

MaVia = MbVfb

Putting values

mv = 4mVfb

Vfb= v/4= 0.25v

Now coefficient of restitution will be

e= (Vfa - Vfb) /( Vib - Via)

= -0.25v/-v

= 0.25

Answer 2

Answer:

0.25

Explanation:

According to conservation of linear momentum,

mv + 0 = 4mv'

=> v' = v/4

So, e = Relative velocity of separation/ Relative velocity of approach

= (v/4)/v

= 1/4

= 0.25