Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=![9.50\times 10^{3} kg](https://tex.z-dn.net/?f=9.50%5Ctimes%2010%5E%7B3%7D%20kg)
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum
![mu+ MU=mv+ MV](https://tex.z-dn.net/?f=mu%2B%20MU%3Dmv%2B%20MV)
Substitute the values
![9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V](https://tex.z-dn.net/?f=9.5%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20380%20i%2B0.150%280%29%3D9.5%5Ctimes%2010%5E%7B-3%7D%20%28250%29j%2B0.150V)
![3.61i=2.375j+0.150V](https://tex.z-dn.net/?f=3.61i%3D2.375j%2B0.150V)
![3.61 i-2.375j=0.150V](https://tex.z-dn.net/?f=3.61%20i-2.375j%3D0.150V)
![V=\frac{1}{0.150}(3.61 i-2.375j)](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B0.150%7D%283.61%20i-2.375j%29)
![V=24.07i-15.83j](https://tex.z-dn.net/?f=V%3D24.07i-15.83j)
Magnitude of velocity of stone
=![\sqrt{(24.07)^2+(-15.83)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%2824.07%29%5E2%2B%28-15.83%29%5E2%7D)
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction
![\theta=tan^{-1}(\frac{y}{x})](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7By%7D%7Bx%7D%29)
=![tan^{-1}(\frac{-15.83}{24.07})](https://tex.z-dn.net/?f=tan%5E%7B-1%7D%28%5Cfrac%7B-15.83%7D%7B24.07%7D%29)
![\theta=tan^{-1}(-0.657)](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28-0.657%29)
=33.3 degree below the horizontal
(b)
Initial kinetic energy
![K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2](https://tex.z-dn.net/?f=K_i%3D%5Cfrac%7B1%7D%7B2%7Dmu%5E2%2B0%3D%5Cfrac%7B1%7D%7B2%7D%289.5%5Ctimes%2010%5E%7B-3%7D%29%28380%29%5E2)
![K_i=685.9 J](https://tex.z-dn.net/?f=K_i%3D685.9%20J)
Final kinetic energy
![K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2](https://tex.z-dn.net/?f=K_f%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%5Cfrac%7B1%7D%7B2%7DMV%5E2)
=![\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%289.5%5Ctimes%2010%5E%7B-3%7D%29%28250%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%280.150%29%2828.81%29%5E2)
![K_f=359.12 J](https://tex.z-dn.net/?f=K_f%3D359.12%20J)
Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.