wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A body of mass 300 g kept at rest breaks into two parts due to internal forces. One part of mass 200 g is found to move at a velocity of 12 m s−1 towards east. The velocity of the other part is:

A
24 m s1 towards west
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
14 m s1 towards east
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
24 m s1 towards east
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
54 m s1 towards south
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 24 m s1 towards west
Let the mass of body be M.
It splits into two parts.
Let the mass of first part be m1=200 g=0.2 kg
mass of second part be m2 = 100 g=0.1 kg
;final velocity of first part be v1 =12 m s1
final velocity of other part be v2.
Since the body is initially at rest, initial momentum is zero.
According to the law of conservation of linear momentum,
Initial momentum = final momentum (when no external force is applied)
In this case, initial momentum =final momentum = 0
0=m1v1+m2v20=0.2×12+0.1×v2
v2=0.2×120.1=24 m s1
The negative sign indicates that it is moving in the opposite direction (west) with a velocity of 24 m s1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conservation of Momentum
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon