Question

A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a mass less string. The height to which block rises is:

• A. $\frac { v ^ { 2 } } { 2 g } \left( \frac { m } { M + m } \right) ^ { 2 }$
• B. $\frac { v ^ { 2 } } { 2 g } \left( \frac { M + m } { m } \right) ^ { 2 }$
• C. $\frac { v ^ { 2 } } { 2 g } \left( \frac { m } { M } \right) ^ { 2 }$
• D. $\frac { v ^ { 2 } } { 2 g } \left( \frac { M } { m } \right) ^ { 2 }$

Answer 1

Answer: (A)

$\frac { v ^ { 2 } } { 2 g } \left( \frac { m } { M + m } \right) ^ { 2 }$

Answer 2

The situation is as shown in the figure.

Let V be velocity of the block-bullet system just after collision. Then by the law of conservation of linear momentum, we get

$\mathrm { mv } = ( \mathrm { m } + \mathrm { M } ) \mathrm { V }$

…..(i)

Let the block rises to a height h.

According to law of conservation of mechanical energy,

We get

$\frac { 1 } { 2 } ( m + M ) v ^ { 2 } = ( m + M ) g h$

$\mathrm { h } = \frac { 1 } { 2 \mathrm {~g} } \left( \frac { \mathrm { mv } } { \mathrm { m } + \mathrm { M } } \right) ^ { 2 }$ (Using (i))

.