Question

15. A visitor to a fairground throws a soft object of mass 0.12 kg at a coconut of mass 0.48 kg. The soft object stops moving when it hits the coconut. In order to dislodge the coconut, it must be made to move at 0.10m/s. What is the minimum speed with which the visitor should throw the soft object in order to dislodge the coconut? ​

Answer 1

`0.40\; {\rm m \cdot s^(-1)}`, assuming that that momentum is conserved and the velocity of the projectile doesn't change during the flight.

Step-by-step explanation:

When an object of mass
`m` is travelling with a velocity of
`v`, the momentum
`p` of that object will be
`p = m\, v`.

Under the assumptions that momentum is conserved, sum of momentum should be the same before and after the impact. In other words:

`m_(c)\, u_(c) + m_(p)\, u_(p) = m_(c)\, v_(c) + m_(p)\, v_(p)`,

Where:

• 
  `m_(c) = 0.48\; {\rm kg}` is the mass of the coconut;
• 
  `u_(c) = 0\; {\rm m\cdot s^(-1)}` is the initial velocity of the coconut before the impact, assuming that the coconut was initially not moving;
• 
  `v_(c) = 0.10\; {\rm m\cdot s^(-1)}` is the velocity of the coconut after the impact;
  
• 
  `m_(p) = 0.12\; {\rm kg}` is the mass of the projectile;
• 
  `u_(p)` is the velocity of the projectile before the impact, which needs to be found;
• 
  `v_(p) = 0\; {\rm m\cdot s^(-1)}` since the projectile stops moving after the impact.

Rearrange this equation to find
`u_(p)`, the velocity of the projectile before the impact:

`\begin{aligned}v_(p) &= (m_(c)\, v_(c) + m_(p)\, v_(p) - m_(c)\, u_(c))/(m_(p)) \\ &= ((0.48)\, (0.10) + (0.12)\, (0))/(0.12)\; {\rm m\cdot s^(-1)} \\ &= 0.40\; {\rm m\cdot s^(-1)}\end{aligned}`.