- #1
samdiah
- 80
- 0
A forensic expert needed to find the velocity of a bullet fired from a gun inorder to predict the trajectory of a bullet. He fired a 5.50g bullet into ballistic pendulum with a bob that had a mass 1.75 kg. The pendulum swung to a height of 12.5 cm above its rest position before dropping back down. What was the velocity of the bullet just before it hit and became embedded in the pendulum bob?
I think I have an understanding of the queation, but just wanted to ensure I am on the right path.
I figured that in the middle kenetic energy E=1/2 mv2 will not be conseved, but momentum is conserved. So if u is the velocity of bullet in start then u=((m+M)v)/m
In the swing to the height of 12.5 cm the potential and kenetic energy will be conserved since both the mass of pendulum and bullet swing up after collision.
so v=sqrt(2gh)
v=sqrt(2*9.8*0.125)
=1.565 m/s
u=((m+M)v)/m
=((5.50+1750)1.565)/5.50
=499.599 m/s
Is this question really this easy because its 4 marks?
Please help. Thank you.
I think I have an understanding of the queation, but just wanted to ensure I am on the right path.
I figured that in the middle kenetic energy E=1/2 mv2 will not be conseved, but momentum is conserved. So if u is the velocity of bullet in start then u=((m+M)v)/m
In the swing to the height of 12.5 cm the potential and kenetic energy will be conserved since both the mass of pendulum and bullet swing up after collision.
so v=sqrt(2gh)
v=sqrt(2*9.8*0.125)
=1.565 m/s
u=((m+M)v)/m
=((5.50+1750)1.565)/5.50
=499.599 m/s
Is this question really this easy because its 4 marks?
Please help. Thank you.