Question

1.3
Two trolleys A and B of mass 100 kg and 200 ka
respectively are kept close with a compressed
spring in between them. When released, they move
away and come to rest after covering distances S.
and S, respectively. If the coefficient of friction is
same for both the trolleys, the ratio S.: S. is
(Neglect time of decompression of spring)
(1) 1:2
(2) 1:4
(3) 1:1
(4) 4:1

Answer 1

Given that,

Mass of trolley A = 100 kg

Mass of trolley B = 200 kg

When released, they move  away and come to rest after covering distances s₁ and s₂.

We need to calculate the ratio of velocity

Using conservation of momentum

$m_{1}v_{1}=m_{2}v_{2}$

$\dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}$

$\dfrac{v_{1}}{v_{2}}=\dfrac{m_{2}}{m_{1}}$....(I)

We need to calculate the ratio of distance s₂ and s₁

Using work energy theorem

$W=K.E$

$F\cdot s=\dfrac{1}{2}mv^2$

Put the value of F

For both case,

$\dfrac{\mu m_{1}gs_{1}}{\mu m_{2}s_{2}}=\dfrac{\dfrac{1}{2}m_{1}v_{1}^2}{\dfrac{1}{2}m_{2}v_{2}^2}$

$\dfrac{s_{1}}{s_{2}}=(\dfrac{v_{1}}{v_{2}})^2$

Put the value from equation (I)

$\dfrac{s_{1}}{s_{2}}=\dfrac{m_{2}}{m_{1}}$

Put the value of mass into the formula

$\dfrac{s_{2}}{s_{1}}=\dfrac{100}{200}$

$\dfrac{s_{2}}{s_{1}}=\dfrac{1}{2}$

Hence, The ratio of distance s₂ and s₁ is 1:2

(1) is correct option.