Example from Physics, 8.3 Elastic and Inelastic Collisions

Calculating Final Velocity in a Two-Dimensional Collision

Suppose the following experiment is performed (Figure 8.11). An object of mass 0.250 kg (m₁) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object of mass 0.400 kg (m₂). The 0.250 kg object emerges from the room at an angle of 45º with its incoming direction. The speed of the 0.250 kg object is originally 2 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.

An illustration shows before and after diagrams of two balls, one moving toward the other. The first ball is labeled m one and the second ball is labeled m two and point four kilograms. In the before diagram, a velocity vector, v one, points from m one to m two. In the after diagram, m one is moving in an upward direction with a velocity vector, v one prime, indicating its direction. It forms an angle of forty-five degrees with the x-axis. The following are also shown: before, p one equal p total; after, prime plus p two prime equals p total; net F equals zero; v two equals zero; v two prime is unknown; angle two is unknown.

Figure 8.11 The incoming object of mass m₁ is scattered by an initially stationary object. Only the stationary object’s mass m₂ is known. By measuring the angle and speed at which the object of mass m₁ emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity 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. We chose the coordinate system so that the initial velocity is parallel to the x-axis, and conservation of momentum along the x- and y-axes applies.

Everything is known in these equations except v′₂ and θ₂, which we need 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
First, we’ll solve both conservation of momentum equations ( $m_{1} v_{1} = m_{1} v^{'}_{1} \cos θ_{1} + m_{2} v^{'}_{2} \cos θ_{2}$ and $0 = m_{1} v^{'}_{1} \sin θ_{1} + m_{2} v^{'}_{2} \sin θ_{2}$ ) for v′₂ sin $θ_{2}$ .

For conservation of momentum along x-axis, let’s substitute sin $θ_{2}$ /tan $θ_{2}$ for cos $θ_{2}$ so that terms may cancel out later on. This comes from rearranging the definition of the trigonometric identity tan $θ$ = sin $θ$ /cos $θ$ . This gives us

$m_{1} v_{1} = m_{1} v^{'}_{1} \cos θ_{1} + m_{2} v^{'}_{2} \frac{\sin θ_{2}}{\tan θ_{2}} .$
8.14

Solving for v′₂ sin $θ_{2}$ yields

$v^{'}_{2} \sin θ_{2} = \frac{(m_{1} v_{1} - m_{1} v^{'}_{1} \cos θ_{1}) (\tan θ_{2})}{m_{2}} .$
8.15

For conservation of momentum along y-axis, solving for v′₂ sin $θ_{2}$ yields
$v^{'}_{2} \sin θ_{2} = \frac{- (m_{1} v^{'}_{1} \sin θ_{1})}{m_{2}} .$
8.16

Since both equations equal v′₂ sin $θ_{2}$ , we can set them equal to one another, yielding

$\frac{(m_{1} v_{1} - m_{1} v^{'}_{1} \cos θ_{1}) (\tan θ_{2})}{m_{2}} = \frac{- (m_{1} v^{'}_{1} \sin θ_{1})}{m_{2}} .$
8.17

Solving this equation for tan $θ_{2}$ , we get

$\tan θ_{2} = \frac{v^{'}_{1} \sin θ_{1}}{v^{'}_{1} \cos θ_{1} - v_{1}} .$
8.18

Entering known values into the previous equation gives

$\tan θ_{2} = \frac{(1.50) (0.707)}{(1.50) (0.707) - 2.00} = - 1.129 .$
8.19

Therefore,

$θ_{2} = \tan^{- 1} (- 1.129) = 312° .$
8.20

Since angles are defined as positive in the counterclockwise direction, m₂ is scattered to the right.

We’ll use the conservation of momentum along the y-axis equation to solve for v′₂.

$v^{'}_{2} = - \frac{m_{1}}{m_{2}} v^{'}_{1} \frac{\sin θ_{1}}{\sin θ_{2}}$
8.21

Entering known values into this equation gives

${v^{'}}_{2} = - \frac{(0.250)}{(0.400)} (1.50) (\frac{0.7071}{- 0.7485}) .$
8.22

Therefore,

${v^{'}}_{2} = 0.886 m/s .$
8.23

0 of 2 steps open

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Strategy

Solution

Strategy

Solution