Question

Two ends of a uniform garland of mass m and length L are hanging vertically as shown. At instant t the end Z is released. If y is the distance moved by this end in time dt, change in momentum of an element of length dy is given as

• A. $\frac { \mathrm { m } } { \mathrm { L } }$ dy(2gy)^1/2 (down ward)
• B. $\frac { \mathrm { m } } { \mathrm { L } }$ dy (2gy)^1/2 (upward)
• C.
  • $\frac { \mathrm { m } } { \mathrm { L } }$ dy dy $\sqrt { \mathrm { gy } }$ (upward)
• D. $\frac { \mathrm { m } } { \mathrm { L } }$ dy(gy)^1/2 (upward)

Answer 1

Answer: (D)

$\frac { \mathrm { m } } { \mathrm { L } }$ dy(gy)^1/2 (upward)

Answer 2

Let XO = OZ = L/2. When end Z goes down by y then velocity of bend

v = $\sqrt { 2 g \frac { y } { 2 } }$

because bend falls through y/2.

= - $\left( \frac { \mathrm { m } } { \mathrm { L } } \mathrm { dy } \right) \mathrm { v }$

= $- \frac { \mathrm { m } } { \mathrm { L } } \mathrm { dy }$ (gy)^1/2 (downward)

= m/L (gy)^1/2 dy (upward)