Answer to Question #143470 in Mechanics | Relativity for tejanshi tejanshi

Question

A 4kg mass is attached to a k=350 N/m spring then compressed 50cm then released(see diagram below). It travels through equilibrium to a point 20 cm right of equilibrium where it makes a e= 0.75 collision with a stationary 2 kg mass. Find

a) the speeds of the masses after the collision

The 4 kg remains on the spring and now moves in SHM.

b) find the amplitude of this SHM

c) find the initial phase of the SHM (t=0 is the instant after collisio

Answer 1

let $u_{1}$ = initial velocity of 4kg mass before collision

$u_{2} =$ velocity of 2kg mas before collision

$v_{1} =$ velocity of 4kg mass after collision

$v_{2}$ = velocity of 2kg mass after collision

$1/2k(0.5)^2-1/2k(0.2)^2 = 1/2m_{1}u^2_{1}$

$350(0.25-0.04)=4u^2_{1}$

$2u_{1}=8.57$ and so $u_{1}=4.285m/s$

$v_{1}= \frac{m_{1} - em_{2}}{m_{1}+m_{2}} u_{1} +\frac{(1+e)m_{2}}{m_{1}+m_{2}}u_{2}$

$v_{2} = \frac{m_{2}-em_{1}}{m_{1}+m_{2}}u_{2}+ \frac{(1+e)m_{1}}{m_{1}+m_{2}}u_{1}$

(a) $m_{1}=4kg, u_{1} = 4.285m/s, m_{2}= 2kg, u_{2}=0$ and $e = 0.75$

$v_{1}= \frac{4-0.75\times2}{4+2} \times 4.285 = 1.785m/s$

$v_{2} = \frac{(1+0.75) \times 4\times4.285}{4+2}= 4.999m/s$

(b) Let A be the amplitude of SHM.

then $1/2kA^2 = 1/2m_{1}v^2_{1}+1/2k(0.2)^2$

$350A^2 = 4(1.785)^2+350(0.2)^2$

$A=0.276m = 27.6cm$

c) $x = Asin\phi$ where $\phi$ is the initial phase.

At $t= 0, x = 0.2m$

$0.2 = 0.276sin\phi$

$sin\phi = 0.725$

$\phi = sin^{-1}(0.725) = 46.46^{0}$