A bullet of mass 20g strikes a pendulum of mass 5kg. The center of mass of the pendulum rises a vertical distance of 10cm. If the bullet gets embedded into the pendulum, calculate the initial speed of the bullet.
A
354.97m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
340.97m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
250.41m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
251m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A354.97m/s Given: Mass of bullet, m1=0.02kg Mass of pendulum, m2=5kg Center of mass of pendulum rises to a height h=0.1m
Let speed of bullet =u1 Pendulum is at rest, so u2=0 Let the common velocity of bullet and pendulum =v v=m1u1+m2u2m1+m2=0.02×u1+5×05+0.02 i.e v=u1251 ... (1)
After bullet gets embedded, applying conservation of energy (m1+m2)gh=12(m1+m2)v2 v=√2gh=√2×10×0.1 v=√2m/s From eq.(1) √2=u1251 or u1=354.97m/s