Question

2.
Two blocks of masses 2 kg and 3 kg are connected
by massless spring and are placed on smooth
surface. If impulse is given to 2 kg block and block
starts moving with velocity 10 m/s in (+ve) X-
direction, then velocity of 2 kg block w.rt centre of
mass is
(1) 4 m/s (+ve) x-direction
(2) 6 m/s (+ve) x-direction
(3) 4 m/s (-x) direction
(4) 6 m/s (-x) direction
uing of mass 'm'​

Answer 1

Answer:

The block velocity will be 6 m/s in +ve x-direction

Step-by-step explanation:

m₁ = 2kg

m₂ = 3kg

v₁ = 10m/s

v₂ = 0m/s

velocity w.r.t center of mass = $v_{cm}$ = $\frac{m 1v1 + m2v2}{m1 +m2}$

 $v_{cm}$ = $\frac{2(10)+3(0)}{2+3}$

$v_{cm}$  = 4 m/s

Now body of 3kg bock with respect to center of mass

= 0 - 4 = 4m/s

And velocity w.r.t centre of mass of 2kg block

= 10 - 4 = 6m/s

The block velocity will be 6 m/s in +ve x-direction.

Hope the answer may help!

Answer 2

Answer:

The block velocity will be 6 m/s in +ve x-direction.

Step-by-step explanation:

In this question,

We have been given that

Two blocks of masses 2 kg and 3 kg are connected by massless spring and are placed on a smooth surface, impulse is given to 2 kg block and its start moving at a speed of 10 m/s.

We need to find the velocity of 2 kg mass w.r.t to centre of mass?

Therefore,

m₁ = 2kg

m₂ = 3kg

v₁ = 10m/s

v₂ = 0m/s

Velocity w.r.t center of mass = $\mathbf{v_{cm}\ =\ \frac{m_{1}v_{1}\ +\ m_{2}v_{2}}{m_{1}\ +\ m_{2}}}$

Putting the values we get,

 $v_{cm}\ =\ \frac{2(10)+3(0)}{2+3}$

$v_{cm}\ =\ 4 m/s$

Now body of 3kg bock with respect to center of mass

= 0 - 4 = 4m/s

And velocity w.r.t centre of mass of 2kg block

= 10 - 4 = 6m/s

The block velocity will be 6 m/s in +ve x-direction.