A block of mass m1 = 1.30 kg moving at v1 = 1.10 m/s undergoes a completely inel

Question

A block of mass m1 = 1.30 kg moving at v1 = 1.10 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.500 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 = 2.80 kg , which is initially at rest. The three blocks then move, stuck together, with speed v3.(Figure 1) Assume that the blocks slide without friction. Part A: Find v2v1, 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. Part B: Find v3v1, 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. Express your answer numerically using three significant figures

Explanation / Answer

here,

mass of Block 1 = M1 = 1.3 kg
mass of Block 2 = M2 = 0.5 kg
mass of Block 3 = M3 = 2.80 kg

Case A:

From Inelastic collision, Momentum Before Collison = Momentum After Collison

M1.V1 = M1V2 + M2V2
V1/V2 = (M1+M2) / M1
V1/V2 =( 1.3 + .5 )/ 1.3
V1/V2 = 1.38

Case B:again
From Inelastic collision, Momentum Before Collison = Momentum After Collison

M1*V1 = M1*V3 + M2*V3 + M3*V3

V3/V1 = M1/(M1+M2+M3)

V3/V1= 1.3 / (0.3+0.5+2.80)

V3/V1= 0.285

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