Q5.3 Making use of the centre of mass I (S)

A projectile of mass 3 kg is fired horizontally from the top of a cliff 100 m high with a speed of 10 ms-1. During its flight, it breaks into two fragments, which reach ground level at the same time. The smaller fragment, of mass 1 kg lands 30 m from the cliff. Where does the other fragment land?
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Hint

Solve the trajectory problem for the center of mass. If we know where the center of mass and one fragment lands, where does the other fragment land?
 

Solution

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Solution

Graphic - No title - commotion1

The motion of the centre of mass of the projectile is unaffected by the breakup, but continues along the original trajectory as shown in the diagram. The x and y co-ordinates of the centre of mass are given by

\[ x_{CM}=ut;\; y_{CM}=h-\frac{1}{2}gt^{2}. \]

Say the fragments hit the ground at time t=T. The centre of mass must reach the ground at the same time. Thus yCM=0 when T=(2h/g)1/2. At this time xCM=uT=u(2h/g)1/2=45.2 m.

Now we have to calculate the center of mass at the bottom of the cliff. Use (Key Point 4.1).

If the fragment of mass m1=1 kg lands at x1=30 m, and the fragment of mass m2=2 kg lands at x2, then

MxCM=(m1+m2)xCM=m1x1+m2x2.

Solving for x2, we obtain

\[x_{2}=\frac{(m_{1}+m_{2})x_{CM}-m_{1}x_{1}}{m_{2}}=52.8\,m.\]