Question

The length of a ballistic pendulum is 1 m and the mass of its block is 0.98 kg. A bullet of mass 20 g strikes the block along horizontal direction and gets embedded in the block. If block + bullet complete vertical circle of radius 1 m, the striking velocity of bullet is:
(A) 280 m/s
(B) 350 m/s
(C) 420 m/s
(D) 490 m/s

Answer 1

Hint: Apply the conservation of momentum; initial momentum of the frame will be equal to final momentum. Also think about the necessary condition for an object to make a complete circle.

Complete step-by-step solution:
We are given that a bullet of mass 20g strikes the block and then the combined body of weight 980g+ 20g completes a full revolution. For a bob to complete a circle, it should
\[{{\text{v}}_{{\text{min}}}}{\text{ = }}\sqrt {{\text{gr}}} \]
This is the velocity at the top end of the circle, similarly we can find the velocity at the bottom of the circle by applying conservation of energy,
\[{{\text{v}}_{{\text{max}}}}{\text{ = }}\sqrt {{\text{5gr}}} \]
This should be the minimum speed of the combined object at the bottom, now applying conservation of momentum,
Initial momentum = final momentum
\[
  {{\text{m}}_{{\text{bullet}}}}{{\text{v}}_{{\text{bullet}}}}{\text{ + }}{{\text{m}}_{{\text{block}}}}{{\text{v}}_{{\text{block}}}}{\text{ = (}}{{\text{m}}_{{\text{bullet}}}}{\text{ + }}{{\text{m}}_{{\text{block}}}}{\text{)}}{{\text{v}}_{{\text{max}}}} \\
  {\text{0}}{\text{.02(}}{{\text{v}}_{{\text{bullet}}}}{\text{) + }}{{\text{m}}_{{\text{block}}}}{\text{(0) = (0}}{\text{.02 + 0}}{\text{.98)}}\sqrt {{\text{5x9}}{\text{.81x1}}} \\
  {\text{0}}{\text{.02}}{{\text{v}}_{{\text{bullet}}}}{\text{ = }}\sqrt {{\text{5x9}}{\text{.81x1}}} \\
  {{\text{v}}_{{\text{bullet}}}}{\text{ = 350m/s}} \\
 \]

Hence, option B is correct answer.

Note: This is the case of an inelastic collision, i.e. when the body gets embedded in another body.