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I have this problem here that goes like this:

Two curling rocks of equal mass, one red and the other yellow, are involved in a perfectly elastic, glancing collision. The yellow rock is initially at rest and is struck by the red rock which is moving at a speed $v_{oi}$ =6 m/s. After the collision, the red rock moves along a direction that makes an angle of θ=20 degrees with its initial direction of motion.

(a) What is the speed of the red rock?

(b) What is the speed of the yellow rock?

(c) What angle does the yellow rock make with the initial direction of motion?

I'm a bit stuck at figuring out how to set up the system. I first use conservation of momentum equation here.

Initial: $v_{yellow x}=0$

$v_{yellow y}=0$

$v_{red x}=6$

$v_{red x}=0$

Final: $v_{yellow x}=-v_{yellow}cos(20)$

$v_{yellow y}=-v_{yellow}sin(20)$

$v_{red x}=v_{red}$

$v_{red xy}=0$

If we plug this in to conservation of momentum for x component we get:

$m_{red}(6)=m_{yellow}(-v_{yellow}cos(20))+m_{red}v_{red}$

For y component we get: $0=m_{yellow}(-v_{yellow}sin(20)$

Is this set up correct? Is my interpretation of the problem sound? Would I just simply proceed by also using conservation of energy here? I'm really unsure as to how I'm setting the problem with so many components and variables to keep track of.

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  • $\begingroup$ $m v_{rxo} = m v_{rxf} + m v_{yxf} = m v_{rf}\cos{20} + m v_{yf}\cos{\theta}$(where the subscripts denote color, component of velocity and initial or final), a similar equation will be needed for the y components except they should subtract to zero. Also since both masses are the same m can be dropped from the calculation whenever you like. $\endgroup$
    – Triatticus
    Jun 23, 2016 at 3:55

1 Answer 1

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Initial condition: $v_{\rm ri} = 6, \theta_{\rm ri}=0^\circ, v_{\rm yi}=0$ with $\theta_{\rm yi}$ irrelevant.

Final condition: $\theta_{\rm rf}=20^\circ$, with $v_{\rm rf}, v_{\rm yf}, \theta_{\rm yf}$ all unknown.

Set up:

  • Conserve momentum along x-axis: $~~v_{\rm ri}\cos\theta_{\rm ri}+v_{\rm yi}\cos\theta_{\rm yi}=v_{\rm rf}\cos \theta_{\rm rf}+v_{\rm yf}\cos\theta_{\rm yf}$

  • Conserve momentum along y-axis: $~~v_{\rm ri}\sin\theta_{\rm ri}+v_{\rm yi}\sin\theta_{\rm yi}=v_{\rm rf} \sin \theta_{\rm rf}+v_{\rm yf}\sin\theta_{\rm yf}$

  • Conserve kinetic energy: $\qquad\qquad~v_{\rm ri}^2+v_{\rm yi}^2=v_{\rm rf}^2+v_{\rm yf}^2$

Get Ready?

Go!

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