A bullet of mass m1 is moving with speed v0 hits a sand bag of mass m2. If the speedof the bullet after passing the sand bag is (v0/3), then the height h up to which the bag rises is (Assume g = acceleration due to gravity)

Question

A bullet of mass m1 is moving with speed v0 hits a sand bag of mass m2. If the speedof the bullet after passing the sand bag is (v0/3), then the height h up to which the bag rises is (Assume g = acceleration due to gravity) (A) h=(1/2 g)((2 m1 v0/3 m2))2 (B) h=(2 m1 v0/3 m2) (C) h=(1/2 g) (D) h=((2 m1 v0/3 m2))2

Tardigrade Admin · Accepted Answer

The given situation is as shown in figure,

As, momentum is conserved in collision.

∴ p_{i} = p_{f}

\Rightarrow m_{1} v_{0} = m_{2} v_{2} + m_{1} \frac{v _{0}}{3}

\Rightarrow v_{2} = \frac{2 m _{1} v _{0}}{3 n _{2}} \dots

(i)
Given that, the bag rises upto height

h

.
So, at this height the kinetic energy is equal to potential energy.
i.e.,

K E = PE

\Rightarrow \frac{1}{2} m_{2} v_{2}^{2} = m_{2} g h

\Rightarrow h = \frac{1}{2 g} v_{2}^{2} = \frac{1}{2 g} (\frac{2 m _{1} v _{0}}{3 m _{2}})^{2}

Q. A bullet of mass $m_{1}$ is moving with speed $v_{0}$ hits a sand bag of mass $m_{2}$ . If the speedof the bullet after passing the sand bag is $\frac{v _{0}}{3}$ , then the height $h$ up to which the bag rises is
(Assume $g =$ acceleration due to gravity)

$h = \frac{1}{2 g} (\frac{2 m _{1} v _{0}}{3 m _{2}})^{2}$

$h = \frac{2 m _{1} v _{0}}{3 m _{2}}$

$h = \frac{1}{2 g}$

$h = (\frac{2 m _{1} v _{0}}{3 m _{2}})^{2}$

Q. A bullet of mass m1​ is moving with speed v0​ hits a sand bag of mass m2​. If the speedof the bullet after passing the sand bag is 3v0​​, then the height h up to which the bag rises is (Assume g= acceleration due to gravity)

h=2g1​(3m2​2m1​v0​​)2

h=3m2​2m1​v0​​

h=2g1​

h=(3m2​2m1​v0​​)2

Q. A bullet of mass $m_{1}$ is moving with speed $v_{0}$ hits a sand bag of mass $m_{2}$ . If the speedof the bullet after passing the sand bag is $\frac{v _{0}}{3}$ , then the height $h$ up to which the bag rises is
(Assume $g =$ acceleration due to gravity)

$h = \frac{1}{2 g} (\frac{2 m _{1} v _{0}}{3 m _{2}})^{2}$

$h = \frac{2 m _{1} v _{0}}{3 m _{2}}$

$h = \frac{1}{2 g}$

$h = (\frac{2 m _{1} v _{0}}{3 m _{2}})^{2}$