Question

A trolley of mass $200\,{\text{kg}}$ moves with a uniform speed of ${\text{36}}\,{\text{km/h}}$ on a frictionless track. A child of mass ${\text{20}}\,{\text{kg}}$ runs on the trolley from one end to the other (${\text{10 m}}$ away) with a speed of ${\text{4 m}}{{\text{s}}^{ - 1}}$ relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer 1

There are two questions asked. First find the final speed of the trolley. For this we will need to apply the law of conservation of linear momentum, find the initial and final momentum of the system and using this find the final velocity of the trolley. Once, the final velocity is obtained then use the formula for distance to find the distance moved by the trolley.

Complete step by step answer:
Given, mass of the trolley, $M = 200\,{\text{kg}}$.Speed of the trolley, $v = {\text{36}}\,{\text{km}}{{\text{h}}^{ - 1}}$.Mass of the child, $m = {\text{20}}\,{\text{kg}}$. Distance ran by the child, $d = {\text{10 m}}$.Relative velocity of the child with respect to the trolley, ${v_r} = {\text{4 m}}{{\text{s}}^{ - 1}}$

We convert the speed of the trolley to the standard unit that is meter per second.

\Rightarrow v= 36 \times \dfrac{{1000}}{{3600}}{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \\\ \Rightarrow v= 10{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ We will apply the law of conservation of linear momentum which says the initial and momentum of the system remains the same. Momentum of a body is given by, $$P = {\text{mass}} \times {\text{velocity}}$$ (i) First we find out the initial momentum of the system Initial momentum of the trolley is , $${P_t} = M \times v$$ (ii) Initial velocity of the child will be the same as that of the trolley. Initial momentum of the child is, $${P_c} = m \times v$$ (iii) Initial momentum of the system will be addition of initial momentum of the trolley and the child. Therefore, initial momentum will be, $${P_i} = {P_t} + {P_c}$$ Using equations (ii) and (iii) in the above equation we get, $${P_i} = Mv + mv$$ $$ \Rightarrow {P_i} = \left( {M + m} \right)v$$ Putting the values of $$M$$, $$m$$ and $$v$$ in the above equation we get, $${P_i} = \left( {200 + 20} \right) \times 10$$ $$ \Rightarrow {P_i} = 2210\,{\text{kgm}}{{\text{s}}^{{\text{ - 1}}}}$$ (iv) Now, we will find the final momentum of the system Let the final momentum of the trolley be $${v_t}$$ The relative velocity of the child with respect to the trolley is given as $${\text{4 m}}{{\text{s}}^{ - 1}}$$. So, final velocity of the child will be, $${v_c} = {v_t} - 4$$ Final momentum of the trolley will be (using equation (i)), $${P'_t} = M{v_t}$$ Final momentum of the child will be (using equation (i)), $${P'_c} = m{v_c}$$ Final momentum of the system will be, $$P' = {P'_t} + {P'_c}$$ Putting the values of $${P'_t}$$ and $${P'_c}$$ we have, $$P' = M{v_t} + m{v_c}$$ Putting the values of $$M$$, $$m$$ and $${v_c}$$ we get $${P_f} = 200{v_t} + 20\left( {{v_t} - 4} \right)$$ $$ \Rightarrow {P_f} = 220{v_t} - 80$$ (v) Applying the law of conservation of linear momentum for the system, we get from equations (iv) and (v) $${P_f} = {P_i}$$ Putting the values of $${P_f}$$ and $${P_i}$$ we get, $$220{v_t} - 80 = 2200$$ $$ \Rightarrow {v_t} = \dfrac{{2200 + 80}}{{220}}$$ $$ \Rightarrow {v_t} = 10.3\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ Therefore, final velocity of the trolley is $$10.3\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ Let $$t$$ be the time taken by the child to run on the trolley from one end to the other which is $$d = {\text{10 m}}$$. The formula for time taken is, $$t = \dfrac{{{\text{distance covered}}}}{{{\text{speed}}}}$$ Here, the time taken by the child is $$t = \dfrac{d}{{{v_r}}}$$ Putting the values of $$d$$ and $${v_r}$$ , we get $$t = \dfrac{{10}}{4} = 2.5\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$ Distance moved by the trolley will be the final speed of the trolley multiplied by the time taken, which is $${d_t} = {v_t} \times t$$ Putting the values of $${v_t}$$ and $$t$$, we get $${d_t} = 10.3 \times 2.5 \\\ \therefore {d_t} = 25.75\,{\text{m}}$$ Therefore, the distance moved by the trolley from the time the child began to run is $$25.75\,{\text{m}}$$. **Hence, the final velocity of the trolley is $$10.3{\text{ m}}{{\text{s}}^{ - 1}}$$ and distance moved by the trolley from the time the child began to run is $$25.75\,{\text{m}}$$.** **Note:** There are four important conservation laws; these are conservation of energy, conservation of linear momentum, conservation of electric charge and conservation of angular momentum. Remember these laws as you may need to apply these laws in such types of questions.