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A piece of wood with a mass of  0.03\ kg is dropped from the top of a  100\ m  high building. At the same time, a ball with a mass of  0.02\ kg is thrown vertically upwards from the ground at a speed of 100\; m/s The balls are driven into the wood. Then the maximum height that the combined system can reach above the top of the building before falling is \left ( g=10\; m/s \right ).

Option: 1

10 m


Option: 2

30 m


Option: 3

20 m 

 


Option: 4

40 m 


Answers (1)

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Suppose the two collide at point P after time t. Time for particles to collide,

t=\frac{d}{v_{rel}}=\frac{100}{100}=\ 1\ second

Speed of wood before impact =gt=10 \; m/s

Speed of bullet before impact 

v-gt=100-10 = 90\; m/s

The before condition was :

\begin{aligned} & 0.03 \mathrm{~kg} \downarrow 10 \mathrm{~m} / \mathrm{s} \\ & 0.02 \mathrm{~kg} \uparrow 90 \mathrm{~m} / \mathrm{s} \end{aligned}

After collision 

\uparrow\ v\ \ 0.05\ kg

Now conservation of linear momentum before and after collision:

\begin{aligned} & -(0.03)(10)+(0.02)(90) \\ & =(0.05) v \\ & \Rightarrow v=30 \mathrm{~m} / \mathrm{s} \end{aligned}

The maximum height reached by the object

\begin{aligned} & h'=\frac{v^2}{2 g} \\ & =\frac{(30)^2}{2(10)}=45 \mathrm{~m} \end{aligned}

Now let h is the hight above ground when two meet before impact

Or, \left ( 100-h \right )=0.5 \; gt^2 = 5

Or, h=95 \; m

Height above tower =h+h'-100= 40 m 

Posted by

Divya Prakash Singh

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