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Collisions Students • Midterm #2 answer key and full solutions are on D2L, as are your scores. • Clicker scores have also been updated. • Your estimated class grade will be added by 5pm today.. • Midterm #2 was clearly challenging. 80 The average score 70 was 59 with a standard deviation 60 50 of 16.5. I try to 40 have a mix of easier and harder 30 questions. It 20 seems like the mix 10 was tilted toward 0 10 20 30 40 50 60 70 80 90 100 the difficult side. Exam scores 1 One of the challenging problems • 33% got this problem right: • Be sure you understand what is going on here. Feel free to discuss with me after class or set up another 2 appointment. One of the challenging problems • 25% got this problem right: • Be sure you understand what is going on here. Feel free to discuss with me after class or set up another appointment. 3 Thinking about momentum What happens when you drop a tennis ball? While falling, gravity is the only force acting so velocity increases in the downward direction (ay = -g). When it hits the floor there is a normal force. What do we know about the normal force? It’s much larger than gravity and upward After bouncing, the velocity is in the opposite direction. What happens to the momentum of the tennis ball? Since momentum is mass times velocity, it is similar to the velocity. Gravity increases momentum and then the normal force changes it from down to up in a very short time. 4 What happens during the collision? So exactly what happens when the tennis ball hits the ground (or a tennis racket)? The collision time is very short The force increases rapidly and then decreases back to zero The force versus time graph might look something like what is shown: F Reality is likely more complicated. t Using momentum ideas we can avoid the details of the forces involved. 5 Force versus time graphs for bouncing balls For tennis ball, collision takes 6/1000 of a second with a maximum force of 85 N compared to a weight of 0.55 N. For golf ball, collision takes 1/1000 of a second with a maximum force of 220 N compared to a weight of 0.45 N 6 Impulse-momentum theorem For a constant force, the impulse is defined as ! ! ! J = Fnet ⋅ Δt = Fnet ⋅ (t2 − t1 ) If we consider a constant interval then !force over a finite time ! ! ! dp Δp nd Newton’s 2 law Fnet = can be written Fnet = dt Δt ! ! and therefore Δp = Fnet ⋅ Δt This gives us the impulse–momentum theorem ! ! ! ! J = Δp = p2 − p1 The change in momentum of a particle over a time interval is given by the impulse of the net force over the time interval. 7 Clicker question 1 Set frequency to BA Consider two carts, of masses m and 2m, at rest on an air track (no friction). You push each cart for 3 s, exerting equal force on each. At the end, the momentum of the light cart is… A. ¼ the momentum of the heavy cart B. ½ the momentum of the heavy cart C. equal to the momentum of the heavy cart D. twice the momentum of the heavy cart E. four times the momentum of the heavy cart By the impulse-momentum theorem, ! the ! change in momentum is given by the impulse which is J = Fnet ⋅ Δt . The same force is applied over the same time, so the same momentum change must occur. 8 Clicker question 2 Set frequency to BA A ball bounces off the !floor as shown. The direction of the impulse of the ball Δp is ... A. Straight up ↑ B. Straight down ↓ C. Straight right → D. Straight left ← ! E. At an angle −p ( ) ! ! ! ! Vector subtraction: Δp = p2 − p1 = Δ p ! The horizontal component of momentum (px) does not change, while the vertical component goes from down to up so the change is up. 1 p2 9 Getting impulse from force vs time graph Can get impulse from the area under the force vs time curve. ! ! J = F ⋅ Δt One rectangle is a small force over F a long time and the other rectangle is a large force over a small time. Both have the same area and so give the same impulse t 10 Conservation of momentum Consider two objects, A and B, which interact with each other but there are no other forces acting By Newton’s 3rd law, the force A exerts on B is equal and opposite to the force B exerts on A. Since the forces are equal and opposite and the time over which the forces act must be equal, the impulse is also equal and opposite. Therefore, the change in momentum of B by the force A exerts on B is equal and opposite to the change in momentum of A by the force B exerts on A 11 Conservation of momentum We can think of the two objects A and B as a system. Forces between A and B can change the momentum of A and B but the net change of momentum of the system (due to these forces) is 0 Conservation of momentum: If there is no net (external) force on a system, the total momentum is constant. So if we know the total momentum of A and B at any point, then we know the total momentum of A and B at any other point even if the individual A and B momenta have changed! 12 Clicker question 3 Set frequency to BA Consider two carts, of masses m and 2m, at rest on an air track and connected by a compressed spring. After releasing the two carts, the spring expands to its equilibrium point and drops straight down. What is the total momentum of the system composed of the two carts after they are released? BEFORE A. 0 B. mv to the left C. mv to the right D. 2mv to the left E. Impossible to tell AFTER The spring simply allows the two carts to act on each other so there is no outside force. Therefore momentum is conserved and is still 0 13