# Physics Exercises/Conservation of Linear Momentum

## Exercises

conservation of linear momentum and recoil velocity=== A motionless catapult was aimed horizontally and fired a 10.0 kg wood block due west at 50 m/s. If the catapult's mass was 250 kg, what is the recoil velocity of the catapult if friction is ignored?

## Solutions

Conservation of linear momentum or recoil velocity=== Conservation of linear momentum is used to solve this problem. The sum of the momentum of the wood block plus the momentum of the catapult must equal the sum of their momentum before the object was fired, which is zero.

M = mass, V = velocity, c = catapult, w = wood block

${\displaystyle M_{c}V_{c}+M_{w}V_{w}=0}$

${\displaystyle (250.kg)V_{c}+(10.0kg)(50.0m/s)=0}$

${\displaystyle V_{c}={\frac {-(10.0kg)(50.0m/s)}{250.0kg}}}$

${\displaystyle V_{c}=-2.00m/s}$

The negative sign indicates that the direction is opposite to the direction of motion of the wood block, which would be east.

## ALTERNATIVE SOLUTION

solution as we know here ,the external force is zero(The force which provide the forword velocity to the wooden block )so the net momentum of the system will remain constant

Let Pi =the initial momentum of the system (catapult + wooden block) So that Pi = (Pc +Pw) initial Here Pc is the momentum of the catapult and Pw is the momentum of the wooden block As initially both of the catapult and wooden block were at rest so that their initial momentum was zero As the external force on the system is zero so initial momentum of the system must be equals to the final momentum Hence Pf = 0 = (Pw +Pc) finally As Pw = 10*50 =500 And Pc = 250* final velocity of catapult (v) So that (500+250V) =0 V= -2 m/s here negative sign show that the final velocity of the catapult will be in the opposite direction of the wooden block