Type of Collision - General Physics - Solved Past Paper, Exams of Physics

Download Type of Collision - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. (25 pts) A ball of mass 0.2 kg has a velocity of 1.5 î m/s. It collides with another object. After the collision, it has a velocity of −0.78 î m/s. Note that this problem is based on problem P8.40 in the text. Many of the answers are identical. a) (5 pts) If the collision takes place over a 0.05 s time interval, what is the average force exerted on the ball? ∆~p = ~Ft ∆~p = (0.2)(−0.78)− (0.2)(1.5) = −0.456 kg · m/s ~F = −0.456 0.05 = −9.12 N The negative result indicates a force to the left. b) (5 pts) The object with which it collided is another ball of mass 0.3 kg moving initially with a velocity −0.4 î m/s. What is the velocity of the second ball after the collision? Nothing has (yet) been mentioned about the type of collision. So you cannot assume that energy is conserved. However, momentum is conserved, so we can at least start there. ~pi = (0.2)(1.5) + (0.3)(−0.4) = 0.18 kg · m/s ~pf = (0.2)(−0.78) + (0.3)v 0.18 = −0.156 + 0.3v v = 1.12 m/s Note that since the velocity is positive, the 0.3 kg object moves to the right after the collision. c) (5 pts) Is this a (perfectly) elastic collision? Simply saying “Yes” is not enough. There is no physical argument that can be used to back up a “yes” answer other than actually computing the initial and final energies: Ki = (1/2)(0.2)(1.5) 2 + (1/2)(0.3)(−0.4)2 = 0.249 J Kf = (1/2)(0.2)(−0.78) 2 + (1/2)(0.3)(1.22)2 = 0.249 J Since both energies are equal, now you can say that it is a prefectly elastic collision. On the other hand, since these are macroscopic objects, you might be able to argue that energy should not be conserved. But it is still safer to just calculate the energies and see what happens. d) (5 pts) What is the velocity of the center of mass of the system, including both balls, before the collision? ~VCM = N∑ i=0 mi~vi N∑ i=0 mi = (0.2)(1.5) + (0.3)(−0.4) 0.5 = 0.36 m/s e) (5 pts) By how much does the velocity of the center of mass of the system change during the collision? Here there is a physical argument you can use to “avoid” having to do calculations. Since there is no net force on the system, the center of mass will not accelerate. If it doesn’t accelerate, the velocity of the center of mass doesn’t change. However, you can still do the calculation: ~VCM = (0.2)(−0.78) + (0.3)(1.22) 0.5 = 0.36 m/s Since this is identical to the answer in (d), the velocity of the center of mass doesn’t change.