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
