The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s².
Answers
Answer:
The answer is 5cm.
Explanation:
Initial velocity of 2 kg block, v1 = 1.0 m/s
Initial velocity of the 4 kg block, v2 = 0
Let the velocity of 2 kg block, just before the collision be u1.
Using the work-energy theorem on the block of 2 kg mass:
The separation between two blocks, s = 16 cm = 0.16 m
∴(1/2)m×u1^2−(1/2)m×(1)^2 = −μ×mg×s
u1 = √(1)2−2×0.20×10×0.16
u1 = 0.6 m/s
Linear momentum is conserved. due to elastic collision.
Let v1, v2 be the velocities of 2 kg and 4 kg blocks just after collision.
According to the law of conservation of linear momentum, we can write:
m1u1+m2u2=m1v1+m2v2
2×0.6+4×0=2v1+4v2
2v1+4v2 = 1.2 .....Equation (1)
For elastic collision:
Velocity of separation (after collision) = Velocity of approach (before collision)
i.e. v1−v2 = +(u1−u2)
=+(0.6−0)
v1−v2 = −0.6 .....Equation (2)
Subtracting equation (2) from (1), we get:
3v2 = 1.2
v2 = 0.4 m/s
v1 = −0.6+0.4 = −0.2 m/s
Let the 2 kg block covers a distance of S1.
Applying work-energy theorem for this block, when it comes to rest:
(1/2)×2×(0)^2 + (1/2)×2×(0.2)^2
= −2×0.2×10×S1
S1 = 1 cm
Let the 4 kg block covers a distance of S2.
Applying work energy principle for this block:
(1/2)×4×(0)^2 − (1/2)×4×(0.4)^2
= −4×0.2×10×S2
2×0.4×0.4 = 4×0.2×10×S2
S2 = 4 cm
Therefore, the distance between the 2 kg and 4 kg block.
S1 + S2 = 1 + 4 = 5 cm
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