Car of masses (301.1±0.1)g has an initial speed of (2.12±0.03) m/s and moves in
ID: 2009311 • Letter: C
Question
Car of masses (301.1±0.1)g has an initial speed of (2.12±0.03) m/s and moves in the positive X direction. Car B of mass (305.3±0.1)g is initially stationary. The two cars collide. After the collision, carA comes to a stop and car B has a speed of (1.97±0.02)m/s in the positive X direction.a) Calculate the total Initial kinetic energy of the two car system with uncertainty.
b) Calculate the total final kinetic energy of the two car system with uncertainty.
c) is this an elastic collision? explain
Explanation / Answer
mass of the car1 is m1 = 301.1±0.1 ginitial speed of car1 is u1 = 2.12±0.03 m/s final speed of the car 1 is v1 = 0 mass of the car2 is m2 = 305.3±0.1 g initial speed of the car2 is u2 = 0 final speed of the car2 is v2 = 1.97±0.02 m/s a) The initial kinetic energy of the system is Ki = 0.5m1u12 (u2 = 0) The uncertainity in speed lie between maximum of 2.12+0.03 m/s to 2.12-0.03 m/s Therefore the uncertainity in square of speed lie between 2.152 = 4.6225 to 2.092 = 4.3681 So the square of the speed with uncertainity is u12 = (4.6225 + 4.3681)/2 ± (4.6225 - 4.3681)/2 u12 = 4.4953±0.1272 Therefore The uncertainity in kinetic energy lie between maximum of 0.6961485 J to 0.65739905 J The uncertainity in kinetic energy lie between maximum of 0.6961485 J to 0.65739905 J The kinetic energy with uncertainity is K = 0.676773775±0.019374725 J b) The final kinetic energy of the system is Kf = 0.5m2v22 (v1 = 0) The uncertainity in speed lie between maximum of 1.97+0.02 m/s to 1.97-0.02 m/s Therefore the uncertainity in square of speed lie between 1.992 = 3.9601 to 1.952 = 3.8025 So the square of the speed with uncertainity is v22 = (3.9601 + 3.8025)/2 ± (3.9601 - 3.8025)/2 v22 = 3.8813±0.0788 Therefore The uncertainity in kinetic energy lie between maximum of 0.6047666715 J to 0.5802615 J The kinetic energy with uncertainity is K = 0.59251408575±0.01225258575 J c) clearly the kinetic energy of the system before and after the collision is not conserved (mot the same) Therefore the collision will be inelastic in nature The final kinetic energy of the system is Kf = 0.5m2v22 (v1 = 0) The uncertainity in speed lie between maximum of 1.97+0.02 m/s to 1.97-0.02 m/s Therefore the uncertainity in square of speed lie between 1.992 = 3.9601 to 1.952 = 3.8025 So the square of the speed with uncertainity is v22 = (3.9601 + 3.8025)/2 ± (3.9601 - 3.8025)/2 v22 = 3.8813±0.0788 Therefore The uncertainity in kinetic energy lie between maximum of 0.6047666715 J to 0.5802615 J The uncertainity in kinetic energy lie between maximum of 0.6047666715 J to 0.5802615 J The kinetic energy with uncertainity is K = 0.59251408575±0.01225258575 J c) clearly the kinetic energy of the system before and after the collision is not conserved (mot the same) Therefore the collision will be inelastic in nature