Determining the Final Velocity of an Unseen Object from the Scattering of Another Object · Physics Playground

Example 8.7 from College Physics, 8.6 Collisions of Point Masses in Two Dimensions

Determining the Final Velocity of an Unseen Object from the Scattering of Another Object

Suppose the following experiment is performed. A 0.250-kg object $(m_{1})$ is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg $(m_{2})$ . The 0.250-kg object emerges from the room at an angle of $45 . 0º$ with its incoming direction.

The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity $({v'}_{2}$ and $θ_{2})$ of the 0.400-kg object after the collision.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.12 is one in which $m_{2}$ is originally at rest and the initial velocity is parallel to the $x$ -axis, so that conservation of momentum along the $x$ - and $y$ -axes is applicable.
Everything is known in these equations except $v_{2}^{'}$ and $θ_{2}$ , which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the $x$ - and $y$ -directions.
Solution
Solving $m_{1} v_{1} = m_{1} v_{1}^{'} cos θ_{1} + m_{2} v_{2}^{'} cos θ_{2}$ for $v_{2}^{'} cos θ_{2}$ and $0 = m_{1} v_{1}^{'} sin θ_{1} + m_{2} v_{2}^{'} sin θ_{2}$ for $v_{2}^{'} sin θ_{2}$ and taking the ratio yields an equation (in which θ₂ is the only unknown quantity. Applying the identity $(tan θ = \frac{sin θ}{cos θ})$ , we obtain:
$tan θ_{2} = \frac{v_{1}^{'} sin θ_{1}}{v_{1}^{'} cos θ_{1} - v_{1}} .$
8.68

Entering known values into the previous equation gives

$tan θ_{2} = \frac{(1.50 m/s) (0.7071)}{(1.50 m/s) (0.7071) - 2 . 00 m/s} = - 1 . 129 .$
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Thus,

$θ_{2} = {tan}^{- 1} (- 1 . 129) = 311 . 5º \approx 312º .$
8.70

Angles are defined as positive in the counter clockwise direction, so this angle indicates that $m_{2}$ is scattered to the right in Figure 8.12, as expected (this angle is in the fourth quadrant). Either equation for the $x$ - or $y$ -axis can now be used to solve for ${v'}_{2}$ , but the latter equation is easiest because it has fewer terms.
${v'}_{2} = - \frac{m_{1}}{m_{2}} v_{1}^{'} \frac{sin θ_{1}}{sin θ_{2}}$
8.71

Entering known values into this equation gives

$v_{2}^{'} = - (\frac{0.250 kg}{0.400 kg}) (1.50 m/s) (\frac{0.7071}{- 0 . 7485}) .$
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Thus,

$v_{2}^{'} = 0.886 m/s .$
8.73
Discussion
It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further.

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Discussion