Question

A 0.100-kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, travel- ing horizontally at 350 m>s, strikes the stone and rebounds hori- zontally at right angles to its original direction with a speed of 250 m>s. (a) Compute the magnitude and direction of the velocity\\

Answer 1

Answer:

Magnitude is 25.8 m/s and direction is $35.5^{o}$

Explanation:

From the law of conservation of linear momentum

$m_{b}v_{ib}+ m_{s}v_{is}= m_{b}v_{fb}+ m_{b}v_{fs}$ where $m_{b}$ and $m_{s}$ are masses of bullet and stones respectively, $v_{ib}$ and $v_{is}$ are the initial velocities of bullet and stone respectively, $v_{fb}$ and $v_{fs}$ are the final velocities of bullet and stone respectively

Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$

$2.1 Kg.m/s(\hat i)=1.5 Kg.m/s(\hat j)+ 0.1*v_{fs}$

$0.1*v_{fs}=2.1 Kg.m/s(\hat i)- 1.5 Kg.m/s(\hat j)$

$v_{fs}=21 Kg.m/s(\hat i)- 15 Kg.m/s(\hat j)$

$|v_{fs}|=\sqrt {(v_{x}^{2}+v_{y}^{2})}$

Substituting 21 for $v_{x}$ and 15 for $v_{y}$

$|v_{fs}|=\sqrt {(21^{2}+15^{2})}=25.8 m/s$

To find direction

$tan\theta=\frac {v_{y}}{v_{x}}$

$\theta=tan^{-1}(\frac {15}{21})=35.5^{o}$

Therefore, magnitude is 25.8 m/s and direction is $35.5^{o}$