A bullet of mass 10, g moving horizontally wi | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l}
\,\,\,\,\,Mass\,of\,bullet,\,m = 10\,g = 0.01kg\
Initial\,speed\,of\,bullet,\,u = 400\,m{s^{ - 1}}\
Mass\,of\,block,\,M = 2\,kg

Vedclass · Accepted Answer

$120$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,Mass\,of\,bullet,\,m = 10\,g = 0.01kg\
Initial\,speed\,of\,bullet,\,u = 400\,m{s^{ - 1}}\
Mass\,of\,block,\,M = 2\,kg\
Lenght\,of\,string\,l = 5\,m\
Speed\,of\,the\,block\,after\,collision = {v_1}\
Speed\,of\,the\,bullet\,on\,emerging\,from\
block,\,v = ?\
{m{Using}}\,energy\,conseravtion\,principel\
for\,the\,block,
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,{(KE + PE)_{{\mathop{m Re}
olimits} ference}} = {\left( {KE + PE} ight)_h}\
 \Rightarrow \,\frac{1}{2}Mv_1^2 = Mgh\,\,\,or,\,\,\,{v_1} = \sqrt {2gh} \
\,\,\,\,\,{v_1} = \sqrt {2 	imes 10 	imes 0.1}  = \sqrt 2 \,m{s^{ - 1}}\
{m{Using}}\,momentum\,conservation\,principle\
for\,block\,and\,bullet\,system,
\end{array}$

$\begin{array}{l}
{\left( {M 	imes 0 + mu} ight)_{Before\,collision}} = {\left( {M 	imes {v_1} + mv} ight)_{After\,collision}}\
 \Rightarrow \,0.01 	imes 400 = 2\sqrt 2  + 0.01 	imes v\
 \Rightarrow v = \frac{{4 - 2\sqrt 2 }}{{0.01}} = 117.15\,m{s^{ - 1}} \approx 120\,m{s^{ - 1}}
\end{array}$

Similar Questions

The position $x$ of a particle moving along $x$-axis at time $(t)$ is given by the equation $t=\sqrt{x}+2$, where $x$ is in metres and $t$ in seconds. the work done by the force in first four seconds is .............. $J$

A simple pendulum of mass $200\, gm$ and length $100\, cm$ is moved aside till the string makes an angle of $60^o$ with the vertical. The kinetic and potential energies of the bob, when the string is inclined at $30^o$ to the vertical, are

A body is projected from ground at an angle ${\theta _1}$ with the horizontal with kinetic energy $K$. What is the potential energy of the body, when the velocity makes an angle ${\theta _2}$ with the horizontal?

A body of mass $m= 10^{-2}$ $kg$ is moving in a medium and experiences a frictional force $F= -kv^2$. Its initial speed is $v_0= 10$ $ms^{-1}$. If, after $10\; s$, its energy is $\frac{1}{8}$ $mv_0^2$ the value of $k$ will be

Given that a force $\vec F$ acts on a body for time $t_1$ and displaces the body by $\vec d$ . In which of the following cases the velocity of the body must increase?