Question

A particle of mass m moving with horizontal speed
6 m/s collides elastically with another particle of mass
M(m<<M) moving in the same direction with speed
4 m/s. Find the speed of lighter particle after collision​

Answer 1

Answer:

Explanation:

Hope it helps

Answer 2

Answer:

2 m/s.

Explanation:

From the question we get that the particles collide elastically, so if we take  the speed of the lighter particle as v1 and the heavier particle as v2. So, according to conservation theorem.

From v1−v2=−u1+u2 we will get that (m+M)v1=(m−M)u1+2Mu2 hence, from here if we take the v1 then we will get   v1 as (m-M/m+M)u1 + u2(2m/m+M). Since, it is given that the m<<M so the value of v1 will be -m/m*6 + 2*4 which on solving we will get 2 m/s.