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

In an elastic collision between two masses m and 3m with the same initial speed v but opposite directions, the final velocity of the heavier mass will be half its initial speed in the same direction it was initially moving (-0.5v).

Step-by-step explanation:

The problem presents a case involving conservation of momentum and elastic collisions. When two masses collide elastically, both momentum and kinetic energy are conserved. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Given that one mass is m and the other is 3m, and they are moving with the same speed v but in opposite directions along the x-axis, the final speeds can be determined using conservation laws.

For the described scenario with the heavier object being three times the mass of the lighter object, the final velocity of the heavier object after an elastic collision can be represented by:

v' = ((m - 3m)/(m + 3m)) * v

v' = (-2m/(4m)) * v

v' = -0.5v

Since the collision is elastic and the masses are moving toward each other with equal magnitude of speed, the heavier mass will simply continue in the same direction but with half the initial speed.

Answer 2

Zero

Step-by-step 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 = (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.