Question

A block of mass m is given a velocity v. Find to what height the block will rise after breaking off from mass M. Assume all surface to be smooth.

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

Answer 1

Answer: (D)

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

Answer 2

mv = (m + M)v_x or v_x = $\frac { \mathrm { mv } } { \mathrm { m } + \mathrm { M } }$

Energy conservation gives

$\frac { 1 } { 2 } \mathrm { mv } ^ { 2 } = \frac { 1 } { 2 } ( \mathrm {~m} + \mathrm { M } ) \mathrm { v } _ { \mathrm { x } } ^ { 2 } + \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { y } } ^ { \prime 2 } + \mathrm { mgh } _ { 1 }$

or $\frac { 1 } { 2 } m ^ { 2 } = \frac { 1 } { 2 } ( m + M ) \left( \frac { m v } { M + m } \right) ^ { 2 } + \frac { 1 } { 2 } m _ { y } ^ { \prime 2 } + m _ { y h }$

or v'_y² = v² - $\frac { \mathrm { mv } ^ { 2 } } { \mathrm {~m} + \mathrm { M } }$ - 2gh

If h₂ is the height from break off point then h₂ = $\frac { \mathrm { v } _ { \mathrm { y } } ^ { \prime 2 } } { 2 \mathrm {~g} }$

and h = h₁ + h₂

Thus h = h₁ + h₂ = $\frac { v ^ { 2 } } { 2 g } - \frac { m v ^ { 2 } } { 2 g ( m + M ) }$