Question

A 6.0 kg object is moving at 5.0 m/s along the x axis in the positive direction. It collides

with and sticks to a 2.0 kg object moving also along the x axis. After the collision the
composite object is moving 2.0 m/s along the x axis in the negative direction.
Determine the velocity (magnitude and direction) of the 2.0 kg object before the collision.

Answer 1

V2 = 23 [m/s] to the left.

Step-by-step explanation:

In order to solve this problem, we must use the definition of conservation of linear momentum. That is, the momentum is conserved before and after the collision. The values before the collision will be taken to the left of the equality, and the values after the collision will be taken to the right of the equality, in this way we have:

Σbefore = Σafter

ΣPbefore = ΣPafter

where:

P = m*v

The positive momentum will be taken to the right and the negative momentum is to the left in this way we formulate the following equation:

`(m_(1)*v_(1)) + (m_(2)*v_(2))=-(m_(1) +m_(2))*v_(3)\\`

where:

m1 = mass of the first object = 6 [kg]

v1 = velocity of the first object = 5 [m/s]

m2 = mass of the stick = 2 [kg]

v2 = velocity of the stick [m/s]

v3 = velocity of the composite object = - 2 [m/s]

(6*5) + (2*V2) = - (6 + 2)*2

30 + (8*2) = - 2*V2

46 = - 2*V2

V2 = - 23 [m/s]

Note: the negative sign means the stick moves to the left