CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A vessel at rest explodes into three pieces. Two pieces having equal masses fly off perpendicular to one another with the same velocity of 30 m/s. The third piece has three times mass of each of the other piece. The magnitude and direction of the velocity of the third piece would be


A

102 m/second and 135 from either

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

102 m/second and 45 from either

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

102 m/second and 135 from either

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

102 m/second and 45 from either

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

The correct option is A

102 m/second and 135 from either


Let's say two pieces are having equal mass 'm' and third piece have a mass of '3m'.

According to law of conservation of linear momentum. Since the initial momentum of the system was zero, therefore final momentum of the system must be zero i.e. the resultant of momentum of two pieces must be equal to the momentum of third piece. We know that if two particle possess same momentum and angle in between them is 90 then resultant will be given by P2=mv2=m302

Let the velocity of mass 3m is V. So 3mV=30m2

V=102 and angle 135 from either.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
A Sticky Situation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon