Two blocks of masses m1-1.84 kg and m2-5.03 kg are each released from rest at a

Question

Two blocks of masses m1-1.84 kg and m2-5.03 kg are each released from rest at a height of h-4.81 m on undergo an elastic head-on collision frictionless track, as shown in the figure below, and m2 Determine the velocity of the m (in m/s) - 1.84 kg block just before the collision. A: 1.75OB: 2.33OC: 3.10OD: 4.13 CE: 5.49 OF: 7.30G: 9.71H: 1.29x101 Submit Answer Tries oy3 Determine the magnitude of the velocity of the m2- 5.03 kg block just after the collision. (in m/s) A: 6.93x101B: 1.00 C: 1.46D:2.11E:3.06OF: 4.44 G:6.44 H: 9.33 Submit Answer Tries 0/3 Determine the magnitude of the velocity of the m1-1.84 kg block just after the collision. (in mys) A 3.30! B 4.501 C S 981 D 796| E: 1.06x 101| F: 1.4 1x 101| G 1.87 x 1011 H 2.49×101 Submit Answer Tries 0/3

Explanation / Answer

given m1 = 1.84 kg

m2 = 5.03 kg

height h = 4.81 m

elastic head on collision

a. just befire collision speed of m1 = u1

from conservatiob of energy

0.5m1u1^2 = m1*gh

0.5*u1^2 = 9.81*4.81

u1 = 9.7145 m/s

b. as u1 does not depend on m1, u2 = 9.7145 m/s

let speeds of the blocks after collision be v1 and v2

then from conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

consider speed to right to be +ve

then

(1.84 - 5.03)*9.7145 = 1.84v1 + 5..03v2 = -30.989255

from coefficient of restitution of elastic collision

(v1 - v2)/(u2 - u1) = 1

v1 - v2 = -19.429

v1 = v2 - 19.429

hence

1.84(v2 - 19.429) + 5..03v2 = -30.989255 = 1.84*v2 - 35.74936 + 5.03v2

v2 = 0.69288 m/s speed of block 2 just after collsiion ( to the right)

c. v1 = v2 - 19.429 = -18.736 m/s

hence

v1 = 18.736 m/s to the left

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