Question

A 4.5 kg object moving in the +x direction at 5.5 m/s collides head-on with a 2.5 kg object moving in the -x direction at 4.0 m/s. Find the final velocity of each mass for each of the following situations. (Take the positive direction to be +x.) (a) The bodies stick together 4.92 X m/s (4.5 kg mass) m/s (2.5 kg mass) (b) The collision is elastic. 4.92 5.92 m/s (4.5 kg mass) m/s (2.5 kg mass) x (c) The 4.5 kg body is at rest after the collision m/s (4.5 kg mass) m/s (2.5 kg mass) d) The 2.5 kg body is at rest after the collision. m/s (4.5 kg mass) m/s (2.5 kg mass) (e) The 4.5 kg body has a velocity of 4.0 m/s in the -x direction after the collision m/s (4.5 kg mass) 5.1 x m/s(2.5 kg mass) Are the results in (c), (d), and (e) reasonable"? Explain. 210 F 34店 Type here to search SHIBA

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Answer 1

If the collision is perfectly completely inelastic, it obeys only conservation of linear momentum. Here, V is the combined velocity of the two masses. Solve the above equation forV m, +7m2 (4.5kg) (5.5 m/s)+(2.5 kg) (-4 m/s) 4.5kg+2.5 kg - 2.10 m/s 2.10 m/s (4.5 kg mass) 2.10 m/s (2.5 kg mass) Answers Let initial velcoity of the first puck is and the initial velcoity of the second puck is u2 Let i and v, are the final velocities of masses m, and m,.respectively.As there is no external force acting on the system, lnear momentum conserved. That is total initial momentum is equal to total final momentum From conservation of energy ml (14-5)(4+4)-n": (v2-14 )(4 +14 ) Divide equation (2) by (1) we get (2) Substitute equation (4) in equation (1)and rearrange for - 1M m +m 2(25kg)40 ( 4 Sig. 2.5kg)55 45kg 5 2 Sig-40) 4.5kg +2.5kg 1.2857m/ s = 1.29ms substitute equation (3) in equation (1) and rearrange for 2m12- 2(4.5kg) =(4.5kg +2.5kg 8.2142m/s 25k45(-4.0m/s) 4.5kg+ 25kg 5.5)+

Answers 1.29 m/s (4.5 kg mass) 8.21 m/s (2.5 kg mass) c) According to the conservation of linear momentum -m,v 1n (4.5 kg (5.5 m/s)+(2.5 kg )(-4 m/s)- (4.5 k) (0m/s) 2.5kg 5.9m/s Answers 0 m/s (4.5 kg mass) 5.9 m/s (2.5 kg mass) d) According to the conservation of linear momentum m1111 + m2112-m.. 17 (4.5 kg (5.5 m/s)+(2.5 kg)(-4 m/s)-(2.5 kg)(0m/s) 4.5kg -3.28m/s Answers 3.28 m/s (4.5 kg mass) 0m/s (2.5 kg mass) According to the conservation of linear momentum 1n (4.5 kg) (5.5 m/s)+ (2.5 kg)(-4m/s)-(4.5 kg)(-4.0m/s) 2.5kg 13.1m/s Answers -4.0 m/s (4.5 kg mass) 13.1m/s (2.5 kg mass