A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If V1 and V2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Question

A spring is compressed between two toy carts of mass m1 and m2. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If V1 and V2 are the velocities of the toy carts and there is no friction between the toy carts and the ground, then : (A) V2 / V2 = m1 / m2 (B) V1 / V1 = m2 / m1 (C) V1 / V2 = - m2 / m1 (D) V1 / V2 = m1 / m2

Tardigrade Admin · Accepted Answer

Here, m1 (d V1/d t)=-m2 (d V2/d t) or m1 a1=m2 a2 ldots (1) now, V1=a1 t and V2=a2 t (by using v=u+ at) from (1), m 1 ( V 1/ t )=- m 2 ( V 2/ t ) ∴ ( V 1/ V 2)=-( m 2/ m 1)

$V_{2} / V_{2} = m_{1} / m_{2}$

$V_{1} / V_{1} = m_{2} / m_{1}$

$V_{1} / V_{2} = - m_{2} / m_{1}$

$V_{1} / V_{2} = m_{1} / m_{2}$

Here, $m_{1} \frac{d V _{1}}{d t} = - m_{2} \frac{d V _{2}}{d t}$
or $m_{1} a_{1} = m_{2} a_{2} \dots$ (1)
now, $V_{1} = a_{1} t$ and $V_{2} = a_{2} t$ (by using $v = u +$ at)
from (1), $m_{1} \frac{V _{1}}{t} = - m_{2} \frac{V _{2}}{t}$
$∴ \frac{V _{1}}{V _{2}} = - \frac{m _{2}}{m _{1}}$

V2​/V2​=m1​/m2​

V1​/V1​=m2​/m1​

V1​/V2​=−m2​/m1​

V1​/V2​=m1​/m2​

Here, m1​dtdV1​​=−m2​dtdV2​​ or m1​a1​=m2​a2​… (1) now, V1​=a1​t and V2​=a2​t (by using v=u+ at) from (1), m1​tV1​​=−m2​tV2​​ ∴V2​V1​​=−m1​m2​​

$V_{2} / V_{2} = m_{1} / m_{2}$

$V_{1} / V_{1} = m_{2} / m_{1}$

$V_{1} / V_{2} = - m_{2} / m_{1}$

$V_{1} / V_{2} = m_{1} / m_{2}$

Here, $m_{1} \frac{d V _{1}}{d t} = - m_{2} \frac{d V _{2}}{d t}$
or $m_{1} a_{1} = m_{2} a_{2} \dots$ (1)
now, $V_{1} = a_{1} t$ and $V_{2} = a_{2} t$ (by using $v = u +$ at)
from (1), $m_{1} \frac{V _{1}}{t} = - m_{2} \frac{V _{2}}{t}$
$∴ \frac{V _{1}}{V _{2}} = - \frac{m _{2}}{m _{1}}$