CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A bomb of mass 12 kg is dropped by a fighter plane moving horizontally with a speed of 100 ms1 from a height of 1 km from the ground. The bomb exploded after 10 s into two pieces of masses in the ratio 1:5. If the small part started moving horizontally with a speed of 600 ms1, the speed of bigger part will be (g=10 ms2)

A
100 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1065 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
120 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1002 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 120 ms1
Given that,
Initial horizontal velocity of bomb ux=100 m/s
Initial vertical velocity of the bomb uy=0 m/s

Vertical velocity of bomb after 10 s will be
vy=uy+gt
vy=gt=100 m/s

m1:m2=1:5
m1+m2=12 kg m1=2 kg and m2=10 kg

Applying conservation of linear momentum in Xdirection
mux=m1vx1+m2vx2
12 kg×100 m/s=2 kg×600 m/s+10 kg vx2
vx2=0

Applying conservation of linear momentum in Y direction,
mvy=m1vy2
12kg×100 m/s=2 kg×0+10 vy2
vy2=120 m/s,v2=v2x2+v2y2=120 m/s
(c)

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