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7. [49.09 Points] Two blocks of masses m1 = 2.00 kg and m: = 3.60 kg are each released from rest at a height of

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7. [49.09 Points] Two blocks of masses m1 = 2.00 kg and m: = 3.60 kg are each released from rest at a height of h = 4.00 m on a frictionless track, as shown in the figure below, and undergo an elastic head on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.) (3) Determine the velocity of each block just before the collision. v1,- = m/s V2; = , , , , 7 m/s (b) Determine the velocity of each block immediately aer the collision. Vlf = m/s v2, = m/s (c) Determine the maximum heights to which m1 and m2 rise after the collision. YIf= m YZf = "1 8. [9.09309 Points] PREVIOUS ANSWERS .. G Iain-EMU webassignnet (c) Find the nal velocity of the object if it is initially moving along the x-axis with a velocity of 1.9 m/s. m/s A rie with a weight of 40 N fires a 3.5-g bullet with a speed of 250 m/s. (a) Find the recoil speed of the rie. m/s (b) If a 675N man holds the rie rmly against his shoulder, nd the recoil speed of the man and rie. ' m/s 6. [-19.09 Points] A railroad car of mass 2.0 x 104 kg moving at 4.50 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? m/s (Vb)\"How much kinetic energy is lost in the collision? J 7. [-19.09 Points] DETAILS MY NOTES mini EOED-riowg new g 7 we Iain-HID Safari File Edit View History Bookmarks Window Help 7 85 8 76% Fri 10:55 PM QE ... E webassign.net C 888 Q Oakland University In PS1100-11300.202210: Exam #1: Chapters 1-5 In Course: PHY-1010-11817 / PHY-1080-12068.202210 W Phy 1010 - W22 - Chapter 6 - PHY 1010 - W22 - Ca.. EXERCISE HINTS: GETTING STARTED | I'M STUCK! Find the final velocity of the two balls if the ball with velocity V2; = -21.0 cm/s has a mass equal to half that of the ball with initial velocity V1; = +34.9 cm/s. (Indicate the direction with the sign of your answer.) Vif = cm/s V2f = cm/s 4. [-/9.09 Points] DETAILS MY NOTES The force shown in the force vs. time diagram in the figure below acts on a 2.1-kg object. F (N) 2 -t (s) 2 3 5 (a) Find the impulse of the force. N . S (b) Find the final velocity of the object if it is initially at rest. m/s (c) Find the final velocity of the object if it is initially moving along the x-axis with a velocity of -1.9 m/s. m/s

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