English, asked by Anonymous, 10 months ago

a block of mass m moving at speed v collides with another block of mass 2 m at rest. the lighter block come to rest after the collision. find the coefficient of restitution ?

Answers

Answered by Anonymous

$\huge\boxed{\pink\star\mathfrak\orange{\large{\underline{\underline{Solution:-}}}} }$

Suppose the second block moves at speed v' towards right after the collision. from the principal of conservation of momentum,

$\large\boxed{\mathfrak\red{mv\:=\:2mv'}}$

$\large\boxed{\mathfrak\red{v'\:=\:v/2}}$

hence, the velocity of separation = v/2 and the velocity of approach = v.

$\bold{by\: definition,}$

$\large\bold{e = \frac{velocity \: of \: sepration}{velocity \: of \: approach} }$

$\large\bold{ = > \frac{ \frac{v}{2} }{v} }$

$\large\bold{ = > \frac{1}{2} }$

________________________________

Anonymous: thankyou ♥

Answered by Sukhpreet85

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=u1 =v

Initial velocity of second block=u2 =0

The lighter block comes to rest after collision.

So,

Final velocity of first block=v1 =0

We know that,

From conservation of momentum

m1u1+m2u2= m1v1+m2v2

mv+0= 0+2mv2

v2= v/2

The coefficient of restitution is

e=v2-v1/u1-u2

e= v/2-0/v-0

e= 1/2

e= 0.5

Hence, the value of e is 0.5

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