Learning Goal:
conservation of momentum and coefficient of restitution.
To analyze an oblique impact using the
When an oblique impact occurs between two smooth particles, the particles move away from each other with
velocity vectors that have unknown directions and unknown magnitudes. If the y axis is within the plane of
contact and the x axis is the line of impact, the impulsive forces of deformation and restitution act only along
the line of impact (the x axis). Momentum of the system is conserved along the line of impact (the x axis):
$$\sum m(v_x)_1 = \sum m(v_x)_2$$
The coefficient of restitution, e, relates the relative-velocity components of the particles along the line of impact
(the x axis):
$$e = \frac{(v_B)_2 - (v_A)_2}{(v_A)_1 - (v_B)_1}$$
The moments of both particles A and B are conserved in the plane of contact (the y axis) because no impulse
acts on either particle in this plane. Therefore, the y component of the velocities before and after the collisions
remains unchanged:
$$(v_y)_1 = (v_y)_2$$
Part A
As shown, tennis ball A rolls off the top of a 10.0 m high wall, falls 4.00 m, and strikes another tennis ball, B, obliquely (Figure 1). Before the collision, tennis ball B has a speed of 10.0 m/s as it moves upward. Each ball's mass is 57.0 g and the collision's
coefficient of restitution is 0.830. (Figure 2) In the figure, $\theta_1 = 30.0^\circ$ and $\theta_2 = 20.0^\circ$. What is $(v_A)_1$, the velocity of tennis ball A, immediately before the collision?
Express your answer numerically in meters per second to three significant figures.
View Available Hint(s)
$$(v_A)_1 = 7.7$$
m/s
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