Two persons of mass and are standing at the two ends A and B respectively, of a trolley of mass M as shown.

When the person standing at 'A' position who was initially at rest jumps from the trolley , towards left with $u_{r e l}$ with respect to the trolley then :

the trolley moves towards right

No worries! We‘ve got your back. Try BYJU‘S free classes today!

the trolley rebounds with velocity

\frac{m_{1} u_{r e l}}{m_{1} + m_{2} + M}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

the centre of mass of the system remains stationary

No worries! We‘ve got your back. Try BYJU‘S free classes today!

all the above

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D all the above
The center of mass is an unique point where the weighted relative mass is distributed which sums to be zero.
This is a point which cause force to be applied for linear acceleration without an angular acceleration.

Since no external force acts on the systeam , centre of mass of the system remains constant.

So centre of mass of the system remains stationary
so option C is correct

SInce centre of mass of system is sationary the trolly moves towards right .
so option A is correct

Let the velocity of trolley be $v_{1}$ towards and the velocity of person A be

$v_{2}$ towards left

Then $u_{(} r e l)$ = $v_{1} + v_{2}$

Since initially they are in rest,So initial momentum is 0

Now final momentum : $(M + m_{2}) v_{1} - m_{1} (v_{2})$

Now by conservation of momentum
$p (i n i t i a l) = p (f i n a l)$

$(M + m_{2}) v_{1} - m_{1} (v_{2}) = 0$

$(M + m_{2}) v_{1} = m_{1} (v_{2})$

$v_{2} = [(M + m_{2}) / m_{1}] v_{1}$

Now u(rel) = $[(M + m_{1} + m_{2}) / m_{1}] v_{1}$

Therefore

the trolley rebounds with velocity $\frac{m_{1} u_{r e l}}{m_{1} + m_{2} + M}$
Option B is also correct.

so all the above are correct

Suggest Corrections

Similar questions

Q. A trolley of mass

M = 150 kg

is at rest over a smooth horizontal surface. Two boys each of mass

m = 50 kg

are standing over the trolley. If they jump from the trolley with a relative velocity

v_{r} = 20 m/s

(relative to the velocity of trolley just after jumping) one after other in same direction. The final recoil speed of trolley is

A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u, the centre of mass of the system (trolley $+$ child) will move with a speed.

Q. A child is standing at one end of a long trolley moving with a speed 'v' on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed 'u', the centre of mass of the system (trolley + child) will move with a speed

Q. A man of mass 60 kg jumps from a trolley of mass 20 kg standing on the smooth surface with absolute velocity 3 m/s. Find the velocity of a trolley and total energy produced by man:

Join BYJU'S Learning Program

Two persons of mass and are standing at the two ends A and B respectively, of a trolley of mass M as shown.When the person standing at 'A' position who was initially at rest jumps from the trolley , towards left with urel with respect to the trolley then :

Two persons of mass and are standing at the two ends A and B respectively, of a trolley of mass M as shown.

When the person standing at 'A' position who was initially at rest jumps from the trolley , towards left with $u_{r e l}$ with respect to the trolley then :