Question

Two particles of masses m and 3m are moving toward each other along the x-axis with the same speed v. They undergo a head- on elastic collision and rebound along the x- axis. Determine the final speed of the heavier object.

Answer 1

Answer:0

Explanation:

Given

two particles of mass m and 3m

They  move towards each other with velocity v

As the collision is Elastic therefore value of coefficient of restitution is 1

$e=1$

$e=\frac{Relative\ speed\ after\ collision}{Relative\ speed\ before\ collision}$

if $u_1$ and $u_2$ are velocity of approach of m and 3 m

and $v_1$ and $v_2$ are velocity after collision respectively

therefore

$v_2-v-1=u_1-u_2$

$v_2-v_1=v-(-v)$

$v_2-v_1=2 v$

conserving momentum

$m\cdot v+3m\cdot (-v)=3m\cdot (v_2)+m\cdot v_1$

$-2mv=3mv_2+m(v_2-2v)$

$4mv_2=0$

$v_2=0$

Therefore velocity of heavier mass is zero

 

Answer 2

Answer:

Zero

Explanation:

m1 = m

m2 = 3m

u1 = v

u2 = - v

Let the final speed of m1 and m2 be v1 and v2 respectively.

Use conservation of momentum

m1 x u1 + 2 x u2 = m1 x v1 + m2 x v2

m x v - 3 m v = m v1 + 3m v2

- 2 mv = m (v1 + 3 v2)

v1 + 3 v2 = - 2v ... (1)

As the collision is perfectly elastic so the coefficient of restitution is 1.

Use Newton's formula

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

v - (- v) = v2 - v1

v2 - v1 = 2 v     ... (2)

Add equation (1) and (2), we get

v2 = 0

v1 = - 2v

So, the heavier body comes into rest after collision.