Two blocks of masses m1 = 1.93 kg and m2 = 3.97 kg are each released from rest a

Question

Two blocks of masses m1 = 1.93 kg and m2 = 3.97 kg are each released from rest at a height of h = 4.90 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. Determine the velocity of the mi (in m/s) 1.93 kg block just before the collision 9.801 B: 1.22x 101|| Tries o/3 A: C: 1.53x10|| D: 1.91x10||| E: 2.39x101|| F: 2.99×101 | G: 3.74x101|| H: 4.67x101 Submit Answer Determine the magnitude of the velocity of the m2 (in m/s) 3.97 kg block just after the collision. A: 1.85) B: 2.1 C: 2.3 D: 2.68| E: 3.0 F: 3.421 G: 3.86 | H: 4.36 Submit Answer Tries o/3 Determine the magnitude of the velocity of the m1 - 1.93 kg block just after the collision. (in m/s) A: 1.781 B: 259| C: 3.75| D: 5.44 | E: 7.881 F: 1.14 x 101 G: 1.66x1 H: 2.40x10- Submit Answer Tries 0/3 Determine the maximum heights to which m2 rises after collision. in m) C: 5.83 x 10.11 D: 7.29x 10-1|| E: 9, 11 x 10-11 1.14| 1.42| H: 1.78 A: 3.73x10-11 B: 4.66 x1 Submit Answer Tries 0/3 F: G: Determine the maximum heights to which mi rises after collision. (in m) B: 8.9 C: Tries 0/3 A: 7.181 1.12 x 101| D: 1.40x10-1 E: 1.75 x 10||| F: 2.19 x10| G: 2.74x10L-H:3.42x10- Submit Answer

Explanation / Answer

(a) velocity of both blocks just before collision = (2gh)1/2

=(2×9.8×4.9)0.5 = 9.8 m/sec

(B) it is a perfect elastic collision so

(V2 - V1)/(u1-u2) = 1

V2 - V1 = 9.8 -(-9.8) = 19.6 m/sec.......(1)

By conservation of momentum :-

m1V1+m2V2 = m1u1 + m2u2

1.93V1+3.97V2 = 1.93×9.8 -3.97×9.8 = -19.992............(2)

Solving equation 1 and 2

V1 = -16.58 m/sec

V2 = 3.02 m/sec

(c)V1 = 16.58 m/sec

(d) maximum height for m2

H2 = V22/2g = (3.02)2/2*9.8 = 0.465m

(e)H1 = V12/2g = (16.58)2/2*9.8 = 14.02m

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