Question

Blocks A and B have a mass of 15 kg and 10 kg, respectively. If A is stationary and B has a velocity of 15 m>s just before collision, and the blocks couple together after impact, determine the maximum compression of the spring.

Answer 1

The image of this question showing the spring constant is missing. I have attached it.

Answer:

maximum compression of the spring = 0.288m

Explanation:

From the question, mA = 15 kg, vA = 0m/s, mB = 10 kg, vB = 15 m/s , K= 10Kn/m = 10000N/m

The blocks couple together after impact.

Now, from conservation of linear momentum,

m1V1 + M2v2 = (m1 + m2)(v)

So, (15x0) + (15x10) = (15+10)v

150 = 25v

Thus, V = 150/25 = 6m/s

Now from energy conservation equation and looking at the values gotten above, we arrive at;

(1/2)(m1+m2)v² = (1/2)kx²

Where x is the maximum compression of the spring.

Thus,

(1/2)(15 + 10)(6²) = (1/2)(10,000)x²

(26 x 36 x 0.5) = 5000x²

414 = 5000x²

Divide both sides by 5000

414/5000 = x²

x² = 0.0828

x = √0.0828

x = 0.288m