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From Mechanics

Theory

Exercises

Conservation of Momentum

Key Results

In all collisions, where no external forces act, momentum will be conserved and we can apply

mAvA+mBvB=mAuA+mBuB

or

mAvA+mBvB=mAuA+mBuB

Example 16.1 A bullet of mass 40 grams is travelling horizontally at 250 ms-1. It hits a wooden trolley that is at rest. The bullet and trolley then move together at 10 ms-1.Assume that the bullet and trolley move along a straight line. Find the mass of the trolley.

Solution Before the collision:

uB=250 and uT=0

After the collision:

vB=vT=10

Also the mass of the bullet should be converted to kg:

mB=004

Using conservation of momentum gives:

mBuB+mTuT=mBvB+mTvT004250+mT0=00410+mT1010=04+10mTmT=1010−04=096 kg

Example 16.2 A van, of mass 2.5 tonnes, drives directly into the back of a stationary car, of mass 1.5 tonnes. The van was travelling at 12 ms-1 and both vehicles move together along a straight line after the collision. Find the speed of the vehicles after the collision.

Solution Before the collision: uV=12 and uC=0

After the collision: vV=vC=v

The masses should be converted to kilograms:

mV=2500 and mC=1500

Using conservation of momentum gives:

mVuV+mCuC=mVvV+mCvC250012+15000=2500v+1500v30000=4000vv=400030000=75 ms-1

Example 16.3 Two particles, A and B of mass m and 3m are moving towards each other with speeds of 4u and u respectively along a straight line. They collide and coalesce. Describe how the motion of each particle changes during the collision.

Solution Before the collision: uA=4u and uB=−u

After the collision: vA=vB=v

Using conservation of momentum gives:

mAuA+mBuB=mAvA+mBvBm4u+3m(−u)=mv+3mvmu=4mvv=mu4mu=4u

Example 16.4

A particle, A, of mass 2 kg has velocity (4i+2j) ms-1 . It collides with a second particle, B, of mass 3 kg and velocity (2i−4j) ms-1 . If the particles coalesce during the collision, find their final velocity.

Solution Before the collision: uA=4i+2j and uB=2i−4j

After the collision: vA=vB=v

The masses are defined: mA=2 and mB=3

Using conservation of momentum gives:

mAuA+mBuB=mAvA+mBvB2(4i+2j)+3(2i−4j)=2v+3v8i+4j+6i−12j=5v14i−8j=5vv=514i−8j=28i−16j

Example 16.5 A car, of mass 1.2 tonnes, is travelling at 15 ms-1, when it is hit by a van, of mass 1.4 tonnes, travelling at right angles to the path of the first car. After the collision the two vehicles move together at an angle of 20 to the original motion of the car. Find the speed of the heavier van just before the collision.

Solution This diagram shows the velocities before the collision.

uC=15i and uV=Uj

This diagram shows the velocity after the collision.

vC=vV=Vcos20i+Vsin20j

Using conservation of momentum gives:

mCuC+mVuV=mCvC+mVvV120015i+1400Uj=2600(Vcos20i+Vsin20j)

Considering the i component gives:

120015=2600Vcos20V=1200152600cos20=18026cos20=736 ms-1

Considering the j component gives:

1400U=260018026cos20sin20U=1400262600180tan20=468 ms-1