Learn how to handle collisions in 2 dimensions...and get better at playing billiards.
How can we solve 2-dimensional collision problems?
In other articles, we have looked at how momentum is conserved in collisions. We have also looked at how kinetic energy is transferred between bodies and converted into other forms of energy. We have applied these principles to simple problems, often in which the motion is constrained in one dimension.
If two objects make a head on collision, they can bounce and move along the same direction they approached from (i.e. only a single dimension). However, if two objects make a glancing collision, they'll move off in two dimensions after the collision (like a glancing collision between two billiard balls).
For a collision where objects will be moving in 2 dimensions (e.g. x and y), the momentum will be conserved in each direction independently (as long as there's no external impulse in that direction).
In other words, the total momentum in the x direction will be the same before and after the collision.
Also, the total momentum in the y direction will be the same before and after the collision.
In solving 2 dimensional collision problems, a good approach usually follows a general procedure:
- Identify all the bodies in the system. Assign clear symbols to each and draw a simple diagram if necessary.
- Write down all the values you know and decide exactly what you need to find out to solve the problem.
- Select a coordinate system. If many of the forces and velocities fall along a particular direction, it is advisable to use this direction as your x or y axis to simplify calculation; even if it makes your axes not parallel to the page in your diagram.
- Identify all the forces acting on each of the bodies in the system. Make sure that all impulse is accounted for, or that you understand where external impulses can be neglected. Remember that conservation of momentum only applies in cases where there is no external impulse. However, conservation of momentum can be applied separately to horizontal and vertical components. Sometimes it is possible to neglect an external impulse if it is not in the direction of interest.
- Write down equations which equate the momentum of the system before and after the collision. Separate equations can be written down for momentum in the x and y directions.
- Solve the resulting equations to determine an expression for the variable(s) you need.
- Substitute in the numbers you know to find the final value. Should this require adding vectors, it is often useful to do this graphically. A vector diagram can be drawn and the method of adding vectors head-to-tail used. Trigonometry can then be used to find the magnitude and direction of all the vectors you need to know.
Billiard ball problem
Figure 1 describes the geometry of a collision of a white and a yellow billiard ball. The yellow ball is initially at rest. The white ball is played in the positive-x direction such that it collides with the yellow ball. The collision causes the yellow ball to move off towards the lower-right pocket at an angle of 28° from the x-axis.
The mass of the yellow ball is 0.15 kg and the white ball is 0.18 kg. A sound recording reveals that the collision happens 0.25 s after the player has struck the white ball. The yellow ball falls into the pocket 0.35 s after the collision.
Exercise 1a: What is the velocity of the white ball after the collision?
Exercise 1b: Is the white ball likely to fall into any of the pockets? If so, how could that be avoided?
Exercise 1c: How much energy is lost to the environment in this collision?
Consider the situation in which a baseball is delivered toward a stationary wooden back-board by a pitching machine. The machine is set to deliver the 0.145 kg ball at 10 m/s. The ball hits the board, making an angle of 45° from the surface of the board. The board is slightly flexible and the collision is inelastic. The ball bounces back at an angle of 40° from the surface of the board as shown in figure 3.
Hint: When a ball bounces off a surface, the impulse responsible for the bounce is always directed normal to the surface.
Exercise 2a: What is the velocity of the ball after the collision?
Exercise 2b: If the ball is in contact with the wall for 0.5 ms, what is the magnitude of the force on the ball due to the wall?
Exercise 2c: How much work was done on the wall by the ball?
Want to join the conversation?
- What does it mean to write 0.72angle180degrees? How do we solve them?
Also, can someone please explain exercise 2b?(12 votes)
- No idea if this is relevant anymore but why not. The answer I got was slightly off by 10N but the process should be similar. Starting off, we need to know what Δpx is and what Δpy is, because to find the force on the ball we can use the impulse-momentum theorem and say that FΔt=mΔv. This means that Δp/Δt is equal to force. The only way we can find the Δp is if we fine Δpx and Δpy and add them together (or subtract correct me if I'm wrong, but in this problem the way doesn't really matter, you'll see why). To find Δpx, we have to use the mΔv formula, but to find Δv, we need to carry out trig function. Vf is 9.23 m/s, which we found in 2a. So we take 9.23 m/s and now have to find the x component in the triangle. And with the triangle being a right triangle, we can simply do 9.23cos40. To find the initial, we would just change the numbers accordingly. 9.23 would become 10 and 40 would be 45. And since the ball is moving right on the plane, the question says that that is positive. So then we would subtract Vf-Vi and then multiply by the mass (mΔv remember) meaning that we should get -6.93x10^-5 or -6.93E-5. For the y component, we would carry out the same task, but instead, we would switch the cos out for a sin. Another change is that since the ball bouncing off is considered in the -y direction, we would make 9.23sin40 to -9.23sin40. And once we subtract ((-9.23sin40)-(10sin45)) and multiply that also by 0.145 kg (mΔv), we get -1.8855. So now (correct me if I'm wrong), but I just subtracted the two and got -1.8855 (see the other number was so small the difference barely changed). Then I divided that by 0.0005s (0.5ms converted to seconds), it became -3780N. Hope this helped!(5 votes)
- In exercise 2c: shouldn't the work be equal to the final kinetic energy minus the initial kinetic energy in stead of otherwise? I thought: W = KE_(f) - KE_(i)(10 votes)
- In exercise 1a the total initial momentum is .72. If the final momentum is equal to the initial momentum and the yellow ball gains a velocity of 2.857 and a momentum of .428 then the white ball's velocity could be .72-.428 divided by .18 which is 1.619m/s instead of 2.2m/s?(5 votes)
- I did the exact same thing as +david rominee and +ham1988 . I don't know the answer. I'll try implementing angles into it and see if I get a different answer. If I do, I will notify everyone.(2 votes)
- I don't get it, isn't the baseball diagram different from the blue diagram in the solution? Isn't the 40 degrees outside of the triangle, not inside as shown in the solution? Wouldn't it be 1.025 / cos50 or 1.025 / sin40, rather than 1.025 / cos40?(5 votes)
- On the diagram the shape and orientation of the green triangle is made in a way that it can perfectly fit in the diagram of the baseball striking the backboard. The arrows help to see the direction of the vector. Now I see that your confusion was on which side of the line to go on, because it's either the 50 degree side or the 40 degree side. To keep it simpler they just used the 40 degree angle and since its on a different axis, the cosine and sine switch, to keep the vectors the same. You should be able to get the right hypotenuse.(0 votes)
- In exercise 1C, why is even there any energy loss? We didn´t consider frictional forces nor sound nor gravity. All of the calculations were based in considering no change in momentum. there shouldn´t be any losses. Can someone please provide with an explanation?(3 votes)
- The ending velocities imply that there must have been some energy loss.. The collision was not 100% elastic. The energy loss would be via heat. Conservation of momentum does not at all imply conservation of kinetic energy.(3 votes)
- In the bouncing ball example, why was 1.45 used in the calculations instead of the given 0.145 kg in the problem?(1 vote)
- Hi! Because 0.145 is the mass of the ball, and we are calculating the momentum. So, 0.145 kg * 10 m/s (initial velocity) gives us 1.45 =)(5 votes)
- Can you give an explanation for exercise 2a? I assumed the horizontal speed remained the same and only verticle speed reduced so that the ball bounces off 40 degrees instead of 45 degrees. I don't understand where 1.025 tan (40) and 1.025/cos (40) come from.(2 votes)
- You are correct, the initial x momentum is 0.145 * 10m/s * cos(45) = 1.025. Then divide that x momentum by cos(40) to get the total momentum (x and y) after the collision. This value, 1.025/cos(40), represents the hypotenuse of the right triangle. Thus the leg representing the y momentum (for part 2b) after the collision is that value times sin(40). sin(40)/cos(40) = tan(40), So it reduces to 1.025 * tan(40).(2 votes)
- For questions 2b and 2c, do we only use the velocity (and momentum) in the x direction for our calculations because the y direction was affected by the momentum lost to the wall in the inelastic collision? If this were an elastic collision and no momentum were lost in the y direction, would we do this using both the x and y directions?(2 votes)
- In the bouncing baseball exercise, why is the momentum in the
vertical direction not conserved?
Is it because the impulse from the board is external?
But it is not an external force, since it originates from the system, right?(2 votes)