Question

A block of mass m1 = 1.10 kg moving at v1 = 1.20 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.900 kg . The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 2.40 kg , which is initially at rest. The three blocks then move, stuck together, with speed v3. Assume that the blocks slide without friction. A block of mass m1 = 1.10 kg moving at v1 = 1.20 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.900 kg . The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 2.40 kg , which is initially at rest. The three blocks then move, stuck together, with speed v3. Assume that the blocks slide without friction. (A) Find v2/v1, the ratio of the velocity v2 of the two-block system after the first collision to the velocity v1 of the block of mass m1 before the collision. (B) Find v3/v1, the ratio of the velocity v3 of the three-block system after the second collision to the velocity v1 of the block of mass m1 before the collisions.

Answer 1

A As collision is in elastic, m, o + m₂.0=(M,+ m2) u. > 1'1 x2 = (1'1+0.9) Un lw= 00 66 mysli - Uz = 0.66 = 0.557 ③ (on, + m2) u + mzoo = (mith 2+m3)%3 = 2 x 0.66 = (2+ 2.9 V3. = vz = "32 orys = 0.3 mys. - 4.4 3.1 = 03 = 0.25 1.2

Answer 2

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