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Solution for (a)

The first ball bounces directly into the wall and exerts a force on it in the + x size 12{+x} {} direction. Therefore the wall exerts a force on the ball in the x size 12{ - x} {} direction. The second ball continues with the same momentum component in the y size 12{y} {} direction, but reverses its x size 12{x} {} -component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum.

These changes mean the change in momentum for both balls is in the x size 12{ - x} {} direction, so the force of the wall on each ball is along the x size 12{ - x} {} direction.

Strategy for (b)

Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball.

Solution for (b)

Let u size 12{u} {} be the speed of each ball before and after collision with the wall, and m size 12{m} {} the mass of each ball. Choose the x size 12{x} {} -axis and y size 12{y} {} -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.

p xi = mu ; p yi = 0 size 12{p rSub { size 8{"xi"} } = ital "mu""; "p rSub { size 8{"yi"} } =0} {}
p xf = mu ; p yf = 0 size 12{p rSub { size 8{"xf"} } = - ital "mu""; "p rSub { size 8{"yf"} } =0} {}

Impulse is the change in momentum vector. Therefore the x -component of impulse is equal to 2 mu and the y size 12{y} {} -component of impulse is equal to zero.

Now consider the change in momentum of the second ball.

p xi = mu cos 30º ; p yi = –mu sin 30º size 12{p rSub { size 8{"xi"} } = ital "mu""cos 30"°"; "p rSub { size 8{"yi"} } = - ital "mu""sin 30"°} {}
p xf = mu cos 30º ; p yf = mu sin 30º size 12{p rSub { size 8{"xf"} } = - ital "mu""cos 30"°"; "p rSub { size 8{"yf"} } = - ital "mu""sin 30"°} {}

It should be noted here that while p x size 12{p rSub { size 8{x} } } {} changes sign after the collision, p y size 12{p rSub { size 8{y} } } {} does not. Therefore the x size 12{x} {} -component of impulse is equal to 2 mu cos 30º size 12{ - 2 ital "mu""cos""30"°} {} and the y size 12{y} {} -component of impulse is equal to zero.

The ratio of the magnitudes of the impulse imparted to the balls is

2 mu 2 mu cos 30º = 2 3 = 1 . 155 . size 12{ { {2 ital "mu"} over {2 ital "mu""cos""30" rSup { size 8{ circ } } } } = { {2} over { sqrt {3} } } =1 "." "155"} {}

Discussion

The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative x size 12{x} {} - direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive x size 12{x} {} -direction.

Our definition of impulse includes an assumption that the force is constant over the time interval Δ t size 12{Δt} {} . Forces are usually not constant . Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force F eff that produces the same result as the corresponding time-varying force. [link] shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times t 1 and t 2 size 12{t rSub { size 8{2} } } {} . That area is equal to the area inside the rectangle bounded by F eff , t 1 , and t 2 . Thus the impulses and their effects are the same for both the actual and effective forces.

Figure is a graph of force, F, versus time, t. Two curves, F actual and F effective, are drawn. F actual is drawn between t sub1 and t sub 2 and it resembles a bell-shaped curve that peaks mid-way between t sub 1 and t sub 2. F effective is a line parallel to the x axis drawn at about fifty five percent of the maximum value of F actual and it extends up to t sub 2.
A graph of force versus time with time along the x size 12{x} {} -axis and force along the y size 12{y} {} -axis for an actual force and an equivalent effective force. The areas under the two curves are equal.

Making connections: take-home investigation—hand movement and impulse

Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why?

Practice Key Terms 2

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Source:  OpenStax, Physics 105: adventures in physics. OpenStax CNX. Dec 02, 2015 Download for free at http://legacy.cnx.org/content/col11916/1.1
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