The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
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