An isolated particle of mass m is moving in a horizontal plane (x-y) along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4 . An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at

Question

An isolated particle of mass m is moving in a horizontal plane (x-y) along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4 . An instant later, the smaller fragment is at y = +15 cm. The larger fragment at this instant is at (A) y=-5 cm (B) y=20 cm (C) y=+5 cm (D) y= - 20 cm

Tardigrade Admin · Accepted Answer

Before explosion, particle was moving along

x

-axis i.e., it has no

y

-component of velocity. Therefore, the centre of mass will not move in

y

-direction or we can say

y_{CM} = 0

.
Now,

y_{CM} = \frac{m _{1} y _{1} + m _{2} y _{2}}{m _{1} + m _{2}}

Therefore,

0 = \frac{( m /4 ) ( + 15 ) + ( 3 m /4 ) ( y )}{( m /4 + 3 m /4 )}

or

y = - 5 c m

$y = - 5 c m$

$y = 20 c m$

$y = + 5 c m$

$y = - 20 c m$

Q. An isolated particle of mass m is moving in a horizontal plane(x−y) along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4 . An instant later, the smaller fragment is at y=+15cm. The larger fragment at this instant is at

y=−5cm

y=20cm

y=+5cm

y=−20cm

Before explosion, particle was moving along x-axis i.e., it has no y-component of velocity. Therefore, the centre of mass will not move in y-direction or we can say yCM​=0. Now, yCM​=m1​+m2​m1​y1​+m2​y2​​ Therefore, 0=(m/4+3m/4)(m/4)(+15)+(3m/4)(y)​ or y=−5cm

$y = - 5 c m$

$y = 20 c m$

$y = + 5 c m$

$y = - 20 c m$