Question

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is moving at 2.0 m/s in a direction opposite to the initial direction of motion of the 6.0-kg object. Determine the speed of the 2.0-kg object before the collision. a. 23 m/s b. 15 m/s c. 11 m/s d. 8.0 m/s e. 7.0 m/s

Answer 1

Answer:

a) 23 m/s

Explanation:

• Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       $p_{o} = p_{f} (1)$

• The initial momentum p₀, can be written as follows:

       $p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

• The final momentum pf, can be written as follows:

        $p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

• Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       $v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

• This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.