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Unformatted text preview: CHAPTER 11 COLLISIONS ActivPhysics can help with these problems: Activities 6.2–6.6 Section 111: Impulse and Collisions Problem 1. What is the impulse associated with a 650N force acting for 80 ms? Solution The impulse for a constant force (from Equation 111) is I = ∫ F dt = (650 N)(0 . 08 s) = 52 N · s, in the direction of the force. Problem 2. A 7.5kg object is moving in the positive x direction at 34 m/s when it undergoes a collision that lasts 0.22 s and leaves it moving in the negative x direction at 51 m/s. Find (a) the impulse and (b) the average impulsive force associated with this collision. Solution (a) Equation 111 gives the impulse, I=Δp= m v f m v i = (7 . 5 kg)( − 51 ˆ ı 34 ˆ ı )(m / s) = − 638 ˆ ı kg · m / s, and (b) Equation 112 gives the average impulsive force, F av =I / Δ t = Δp / Δ t = ( − 638 ˆ ı kg · m / s) / (0 . 22 s) = − 2 . 90 ˆ ı kN . Problem 3. A 62kg parachutist hits the ground moving at 35 km/h and comes to a stop in 140 ms. Find the average impulsive force on the chutist, and compare with the chutist’s weight. Solution The average impulsive force (from Equations 111 and 2) is F av =Δp / Δ t = (0 − m v i ) / Δ t = − (62 kg) × (35 m / 3 . 6 s) / (0 . 14 s) = − 4 . 31 kN (opposite to the direction of v i , or upward for most jumps under calm wind conditions). The magnitude is slightly over seven times the parachutist’s weight, or 7.09 mg . Problem 4. (a) By how much does the momentum of a car change when it undergoes a collision during which an average impulsive force of 1 . 2 × 10 5 N acts for 0.25 s? (b) If the car’s mass is 1800 kg, for what initial speed would this collision bring the car to a stop? Solution (a) The momentum changes by Δ p = F av Δ t = (1 . 2 × 10 5 N)(0 . 25 s) = 3 × 10 4 kg · m / s in the direction of the average impulsive force. (b) If p f = 0, Δ p = − p i , and v i = − (3 × 10 4 kg · m / s) / (1800 kg) = − 16 . 7 m / s = − 60 . 0 km / h (opposite to the direction of F av ). Problem 5. A 240g ball is moving with velocity v i = 6 . 7 ˆ ı m / s when it undergoes a collision lasting 52 ms. After the collision its velocity is v f = − 4 . 3 ˆ ı+ 3 . 1 ˆ m / s. Find (a) the impulse and (b) the average impulsive force associated with this collision. Solution (a) From Equation 111, I=Δp = m ( v f v i ) = (0 . 24 kg)( − 4 . 3 ˆ ı+ 3 . 1 ˆ  6 . 7 ˆ ı )(m / s) = ( − 2 . 64 ˆ ı+ . 744 ˆ ) N · s . (b) From Equation 112, F av = I / (0 . 052 s) = ( − 50 . 8 ˆ ı+ 14 . 3 ˆ ) N . Problem 6. A proton moving in the positive x direction at 4 . 3 Mm / s collides with a nucleus. The collision lasts 0.12 fs, and the average impulsive force is 42 ˆ ı+ 17 ˆ μ N . (a) Find the velocity of the proton after the collision. (b) Through what angle has the proton’s motion been deflected?...
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This note was uploaded on 10/31/2010 for the course PHYSICS 69054 taught by Professor Huang during the Fall '09 term at Berkeley.
 Fall '09
 Huang

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