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

A cannon ball is fired with a velocity 200 m/sec at an angle of 60 with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity


A

100 m/s in the horizontal direction

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

300 m/s in the horizontal direction

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

300 m/s in a direction making an angle of 60 with the horizontal

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

200 m/s in a direction making an angle of 60 with the horizontal

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

The correct option is B

300 m/s in the horizontal direction


Momentum of ball (mass m) before explosion at the highest point =mv^i=mu cos 60^i

=m×200×12^i=100 m^i kgms1

Let the velocity of third part after explosion is V

After explosion momentum of system = =P1+P2+P3

=m3×100^jm3×100^j+m3×V^i

By comparing momentum of system before and after the explosion

m3×100^jm3×100^j+m3V^i=100 m^iV=300 m/s


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Cut Shots in Carrom
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon