A block of 23 N in weight rests on a horizontal plane, a sphere of 4.6 N of weight is next to the block in the form of pendulum that hangs of a rope. The sphere rises a height of 2 m from its lower position, is released and collides with the block, the coefficient of restitution between the sphere and the block is 1.
The coefficient of friction between the block and the plane is 0.2.
Determine how fast the sphere will move after the crash? The time in which
Stops the block.
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Let's see. If the coefficient of restitution is 1, the shock is perfectly elastic.
The linear momentum of the system is conserved.
The relative velocity between the bodies before the collision is equal and opposite that after the collision (the kinetic energy is conserved)
We need the rate of fall of the pendular mass.
Vo = √ (2 g h) = √ (2. 9.80 m / s² 2 m) = 6.26 m / s
M = 23 N / 9.80 m / s² = 2.35 kg (block)
M = 4.6 N / 9.80 m / s² = 0.47 kg (sphere)
1) 0.47 kg. 6.26 m / s = 0.47 kg V + 2.35 kg U (1)
V and U are the velocities of the sphere and the block, respectively
From the relation of relative velocities:
6.26 m / s = - (V-U) (2)
Between (1) and (2) there is a system of two equations with two unknowns, which I solve directly.
V = -4.17 m / s (sphere bounces)
U = 2.087 m / s (block velocity immediately after impact)
It is shown that the deceleration on a horizontal rough plane is:
A = u g = 0.2. 9.80 m / s² = 1.96 m / s²
Then Vf = U - a t = 0; (it stops); so that:
T = U / a = 2.087 m / s / 1.96 m / s² = 1.06 seconds (time to stop)
The linear momentum of the system is conserved.
The relative velocity between the bodies before the collision is equal and opposite that after the collision (the kinetic energy is conserved)
We need the rate of fall of the pendular mass.
Vo = √ (2 g h) = √ (2. 9.80 m / s² 2 m) = 6.26 m / s
M = 23 N / 9.80 m / s² = 2.35 kg (block)
M = 4.6 N / 9.80 m / s² = 0.47 kg (sphere)
1) 0.47 kg. 6.26 m / s = 0.47 kg V + 2.35 kg U (1)
V and U are the velocities of the sphere and the block, respectively
From the relation of relative velocities:
6.26 m / s = - (V-U) (2)
Between (1) and (2) there is a system of two equations with two unknowns, which I solve directly.
V = -4.17 m / s (sphere bounces)
U = 2.087 m / s (block velocity immediately after impact)
It is shown that the deceleration on a horizontal rough plane is:
A = u g = 0.2. 9.80 m / s² = 1.96 m / s²
Then Vf = U - a t = 0; (it stops); so that:
T = U / a = 2.087 m / s / 1.96 m / s² = 1.06 seconds (time to stop)
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