Question

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 prefectly elastic.find the separation the blocks is perfectly elastic.find the separation between the two blocks when they come to rest .take g =10m/s².​

Answer 1

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)

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