Question

A girl riding a bicycle along a straight road with a speed of 5 \[m{s^{ - 1}}\] throws a stone of mass 0.5 kg which has a speed of 15 \[m{s^{ - 1}}\] with respect to the ground along with her direction of motion. The mass of a girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is a change in speed, if so?

Answer 1

Hint: In this question we start by writing the initial momentum of the system that is ${P_i} = \left( {{m_g} + {m_s}} \right) \times u$ and the final momentum of the system that is \[{P_f} = {m_g}u + {m_s}{v_s}\]. Now as there is no external force applied on the system so the final momentum is equal to the initial momentum that is $ \Rightarrow {P_i} = {P_f}$. With this we can find the speed of the girl after she throws the stone. Then we can find the change in speed as \[ \Rightarrow \Delta v ={\text{ Speed of girl before throwing stone}}-{\text{speed of girl after throwing the stone }}\].

Complete Step-by-Step solution:
We know that momentum is the quantity of motion a body or a system has.
Let the initial momentum of the system by consisting of girl and stone be ${P_i}$. So we get
${P_i} = \left( {{m_g} + {m_s}} \right) \times u$
Where the mass of girls be ${m_g}$ equal to 50 Kg
             Mass of stone be ${m_s}$ is equal to 0.5 Kg
            Initial velocity be $u$ is equal to 5\[m{s^{ - 1}}\]
Therefore we get
$ \Rightarrow {P_i} = \left( {50 + 0.5} \right) \times 5$
$ \Rightarrow {P_i} = 252.5$\[{m^2}{s^{ - 1}}\]
Now we will find the final momentum ${P_f}$ of the system after the stone is thrown with a velocity ${v_s}$ equal to 15 \[m{s^{ - 1}}\] that is:
\[{P_f} = {m_g}u + {m_s}{v_s}\]
\[ \Rightarrow {P_f} = 50 \times {v_g} + 0.5 \times 15\]
\[ \Rightarrow {P_f} = 50 \times {v_g} + 7.5\] \[{m^2}{s^{ - 1}}\]
Here ${v_g}$ is the velocity of the girl after she throws the stone.
Now we know that as there is no external force applied on the system so the final momentum is equal to the initial momentum
$ \Rightarrow {P_i} = {P_f}$
\[ \Rightarrow 252.5 = 50 \times {v_g} + 7.5\]
\[ \Rightarrow 252.5 - 7.5 = 50 \times {v_g}\]
\[ \Rightarrow 245 = 50 \times {v_g}\]
\[ \Rightarrow {v_g} = \dfrac{{245}}{{50}} = 4.9m{s^{ - 1}}\]
So, yes the speed of the girl changes after she throws the stone and it is equal to 4.9 \[m{s^{ - 1}}\]
Now the change in the speed is equal to
 \[ \Rightarrow \Delta v ={\text{ Speed of girl before throwing stone}}-{\text{speed of girl after throwing the stone }}\]
\[ \Rightarrow \Delta v = 5m{s^{ - 1}}{\text{ }}-{\text{ 4}}{\text{.9}}m{s^{ - 1}}\]
\[ \Rightarrow \Delta v = 0.1m{s^{ - 1}}\]

Note: In these types of questions we use the concept that the final momentum is the same as the initial momentum. But this concept is only used when the external force on an isolated system is zero. Also, we need to be care full with the sign convention that because momentum is a vector quantity.