Question

On a frictionless track, a trolley moves with a speed of 36 km/h with a mass of 200 Kg. A child whose mass is 20 kg runs on the trolley with a speed of 4 m s1 from one end to other which is 20 m. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the  trolley moved from the time the child began to run.
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Answer 1

Question :

On a frictionless track, a trolley moves with a speed of 36 km/h with a mass of 200 Kg. A child whose mass is 20 kg runs on the trolley with a speed of 4 m s1 from one end to other which is 20 m. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the  trolley moved from the time the child began to run.

Given :

Velocity of trolly ( $\tt {v}_{t}$ ) = 36 km/hr

Mass of trolly ($\tt {m}_{t}$) = 200 kg

mass of child ($\tt {m}_{c}$) = 20 kg

Length of trolly (s) = 20 m

Velocity of man relative to trolly ($\tt {v}_{rel}$) = 4 m/s

To Find :

i.Final speed of trolly

ii.Distance the  trolley moved from the time the child began to run.

Formulas :

● p = mv

Where,

p : momentum

m : mass

v : velocity

● v = $\tt \frac{s}{t}$

Where ,

v : velocity

s : displacement

t : time

Solution :

Before running

$\tt {v}_{t}$ = 36 km/hr

$\tt \implies {v}_{t}$ = 36 × $\tt \frac{5}{18}$

$\tt \implies {v}_{t}$ = 2 × 5

$\tt \implies {v}_{t}$ = 10 m/s

$\tt {v}_{rel}$ = 0

$\tt \implies {v}_{m} - {v}_{t} = 0$

$\tt \implies {v}_{m} = {v}_{t}$ = 10 m/s

Initial Momentum = $\tt {p}_{i}$

$\tt \implies {p}_{i} = {m}_{t}{v}_{t} + {m}_{c}{v}_{c}$

$\tt \implies {p}_{i}$ = 200 × 10 + 20×10

$\tt \implies {p}_{i}$ = 2000 + 200

$\tt \implies {p}_{i}$ = 2200 kg m/s

After Running :

Let velocity of trolly be $\tt {v}_{{t}^{'}}$

Velocity of child relative to trolly = $\tt {v}_{rel}$

$\tt \implies {v}_{rel} = {v}_{m} - {v}_{t}$

$\tt \implies - 4 = {v}_{m} - {v}_{{t}^{'}}$

$\tt \implies {v}_{m} = {v}_{{t}^{'}} - 4$

Final momentum = $\tt {p}_{f}$

$\tt \implies {p}_{f} = 200{v}_{{t}^{'}} + 20({v}_{{t}^{'}} -4)$

$\tt \implies {p}_{f} = 220{v}_{{t}^{'}} - 80$

From conservation of momentum

$\tt \implies {p}_{i} = {p}_{f}$

$\tt \implies 220{v}_{{t}^{'}} - 80 = 2200$

$\tt \implies 220{v}_{{t}^{'}} = 2280$

$\tt \implies {v}_{{t}^{'}} = \frac{2280}{220}$

$\tt \implies {v}_{{t}^{'}} = 10.36 \:m/s$

So the final speed of trolly is 10.36 m/s

Time taken by the child to reach other end = t

$\tt \implies$ t = $\tt \frac{s}{v}$

$\tt \implies$ t = $\tt \frac{20}{4}$

$\tt \implies$ t = 5 s

Distance travelled by trolly = $\tt {v}_{{t}^{'}}t$

$\implies$ Distance travelled by trolly = 10.36 × 5

$\implies$ Distance travelled by trolly = 51.8 m

Distance travelled by trolly is 51.8 m