Question

7. The length of a ballistic pendulum is 1 m and
mass of its block is 0.98 kg. A bullet of mass 20
gram strikes the block along horizontal
direction and gets embedded in the block. If
block + bullet completes vertical circle of radius
1m, then the striking velocity of bullet is
1) 280m/s 2) 350m/s 3) 420m/s 4) 490m/s​

Answer 1

Answer:

1/2 (m + M) Vb^2 = 1/2 (m + M) Vt^2 + 2 (M + m) g R

Where Vb is speed of block and bullet at bottom and Vt the speed at top using conservation of energy

Vb^2 = Vt^2 + 4 g R

Vt^2 / R = g    centripetal acceleration at top just equals g

Vb^2 = R g + 4 R g

Vb^2 = 5 R g       Vb = (5 R g)^1/2

m v = (M + m) Vb    conservation of energy of block and bullet

v = (M + m) / m * (5 R g)^1/2

v = 1 / .02 * 7 = 350 m/s

Answer 2

therefore the correct answer is 350m/s