Example two carts on a frictionless track move

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Example: Two carts on a frictionless track move towards each other. Cart A of mass 200 g moves right at 1.5 m/s, cart B of mass 400 g moves left at 1.0 m/s. They stick as a result of the collision. What is the velocity of the stuck carts after they collide? V fx = m A v A,ix + m B v B,ix m A + m B = (200)(1 . 5) + (400)( - 1 . 0) 200 + 400 m/s = - 0 . 16 m/s The stuck carts move left with speed 0.16 m/s.
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Phys 2A - Mechanics Example: Two identical cars approach an intersection. Car 1 is moving east at 35 mph, while car 2 is going north at 55 mph. They collide and stick. What is the velocity of the mangled mess immediately after the collision? ANS: We’d like to use momentum conservation. There are external forces In particular there is friction with the pavement. However, kinetic friction is a small force compared to the collision forces, and we want the velocity “immediately after the collision.” We can neglect the force of kinetic friction during the collision and during the small interval between the collision and our report of the final velocity. Hence: use conservation of momentum, P f = p 1 ,i + p 2 ,i V f = mv 1 ,i + mv 2 ,i m + m = v 1 ,i + v 2 ,i 2 M f V f = m 1 v 1 ,i + m 2 v 2 ,i
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Phys 2A - Mechanics So we need to compute V f = v 1 ,i + v 2 ,i 2 x y v 1 ,i v 2 ,i v 1 ,i + v 2 ,i v 1 ,i = (35 , 0) v 2 ,i = (0 , 55) v 1 ,i + v 2 ,i = (35 , 55) V f = 1 2 (35 , 55) = (17 . 5 , 27 . 5) Compute (all quantities in mph) So the speed immediately after the collision is | V f | = p (17 . 5) 2 + (27 . 5) 2 = 33 mph Direction: = arctan 55 35 = 58 north of east
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Phys 2A - Mechanics Extended version of same Example: Two identical cars approach an intersection. Car 1 is moving north at 35 mph, while car 2 is going east at 55 mph. They collide and stick. If the coefficient of kinetic friction of the cars with the pavement is 0.9, how far do the mangled cars move after the collision? ANS: After determining that V f = 33 mph we use (1) Newton’s Laws to determine acceleration, and (2) kinematics to find displacement y N W x F net f k a = F net m a x = ( f k ) x m = ( - μ k N ) m 0 = N y + W y = N - mg { a x = - μ k g (1) (2) and using v 2 x - v 2 0 x = 2 a x ( x - x 0 ) v x = 0 , v 0 x = V f , a x = - μ k g Δ x = V 2 f 2 μ k g = (33 1609 3600 ) 2 2(0 . 9)(9 . 8) m = 12 m
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Phys 2A - Mechanics Example: Explosions .
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