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Grade: 12th pass
        
  A block of mass m1 = 1.60 kg moving to the right with a speed of 4.00 m/s on a frictionless horizontal track collides with a spring attached to a second block of mass m2 = 2.10 kg moving to the left with a speed of 2.50 m/s, as in Figure 2. The spring has a spring constant of 600 N/m. At the instant when m1 is moving to the right with a speed of 3.00 m/s, determine (a) the velocity of m2 and (b) the distance x that the spring is compressed.
one year ago

Answers : (1)

Arun
24459 Points
							
a) Applying the conservation of momentum, 
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2' 
m1 * (v1 - v1') + m2 * v2 = m2 * v2' 
v2' = m1/m2 * (v1 - v1') + v2 
Let the direction to the right be +, the direction to the left be -. 
So v1 = 4, v2 = -2.5, v1' = 3 
Therefore, 
v2' = 1.6/2.1 (4-3) - 2.5 = -1.74 
Therefore, block2 is moving to the left with a velocity of 1.74 m/s 
b) Applying the conservation of energy, some of the energy after the collision must have transferred to the spring. 
The energy stored in the spring is E = 1/2 * kx^2 
Here, k is the spring constant. 
1/2 * m1 * v1^2 + 1/2 * m2 * v2^2 - 1/2 * m1 * v1'^2 - 1/2 * m2 * v2'^2 = 1/2 * kx^2 
m1 (v1^2 - v1'^2) + m2 (v2^2 - v2'^2) = kx^2 
1.6 * (4^2 - 3^2) + 2.1 (2.5^2 - 1.74^2) = 11.2 + 3.22 = 600 * x^2 
14.42 / 600 = x^2 
x = 0.16 m
one year ago
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