280 eHAPTER 9 Momentum A 600 g air-track glider collides with a spring at one en

Question

280 eHAPTER 9 Momentum A 600 g air-track glider collides with a spring at one end of around i the track. Figure P9.39 shows the glider's velocity and the force exerted on the glider by the spring. How long is the glider in of water backw 3.0 m/s. If other ignored, what is contact with the spring? 47. Ill The flowers BI0 force and speed flower in a mea the acceleratio len grain with 48. I a. With w (s) fle BD b, suppos. FIGURE P9.39 men of the coll Far in space, where gravity is negligible, a 425 kg rocket traveling at 75.0 m/s in the positive x-direction fires its engines. Figure P9.40 shows the thrust force as a function of time. The mass lost by the rocket during these 30.0 s is negligible. a. What impulse does the engine impart to the rocket? b. At what time does the rocket reach its maximum s 40 49. A tennis 10 m/s. She speed of 20 a. How fas You can the brie b. If the te the ave speed? What is the maximum speed? F, (N) 50. I A 20 g 1.0 kg ble and stick a. What b. Use t 1000 r(s) 0 r (ms) 0 10 20 30 5 ms mpu c. Use t the b FIGURE P9.40 FIGURE P9.41 41) lI A 200 g ball is dropped from a height of 2.0 m, bounces o NI a hard floor, and rebounds to a height of 1.5 m. Figure P9.41 d. Doe 51. I Dan shows the impulse received from the floor. What maximum force does the floor exert on the ball? backw 8.0 m/ (42.? A 200 g ball is dropped from a height of 2.0 m and bounces NI on a hard floor. The force on the ball from the floor is shown in 52.1 Jan mass Figure P9.42. How high does the ball rebound? whee same whil 2400 1800 1200 600 spee the r t (ms) 0 0.2 0.4 0.6 FIGURE P9.42 b. FIGURE P9.43

Explanation / Answer

39) The impulse is given by J = m?V = F?t

m?V = F?t

?t = m?V/F = 0.6 x (3-(-3))/(36) = 0.6 x 6/36 = 0.1 second

hence it is in contact for 0.1 second.

40)

The impulse is calculated by finding out the area under the F-t curve

so Area = 0.5 x 1000 x 10+ 0.5 x 1000 x 20 = 15000 Ns

Hence impulse = 15000 N.s

b) The impluse = change in momentum = m?V

m?V = 15000

?V = 15000/m = 15000/425 = 35.3 m/s

Hence the maximum change is of +35.3 m/s

Hence the maximum velocity = 75 + maximum change = 75 + 35.3 = 110.3 m/s

Hence the maximum velocity = 110.3 m/s

it attains maximum velocity when the engine stops delivering so it is at time 30 seconds

hence maximum speed is attained at time = 30seconds

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