Physics, 05.05.2020 18:43, gafran

Two identical blocks with mass 5.0 kg each are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity 2 m/s to the right. The spring is then released by remote control. At some later instant, the left block is moving at 1 m/s to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant

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answered: ajajjsisi9x8

The velocity of the center of mass of the system at that instant is 2m/s.

Given data:

The mass of two identical blocks are, m = 5.0 kg.

The velocity of blocks towards the right is, v = 2 m/s.

The given problem is based on the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.

Total momentum after the collision is,

$p_{t}=(m_{1}+m_{2})v\\\\p_{t}=(5+5) \times 2\\\\p_{t} = 20 \;\rm kg-m/s$

After the spring is released, both objects move remotely with different velocity and have different momentum.

For the block with velocity of 1m/s, its momentum is given as:

$p=m \times v'\\\\p=5 \times -1\\\\p = -5 \;\rm kg.m/s$

For the block moving to the right, its momentum is given as:

$p'=m \times u\\\\p'=5 \times u\\\\p'=5u$

Here u is the velocity of the second block while moving remotely.

Then as per the conservation of momentum,

$p_{t}=p+p'\\\\20 = -5+5u\\\\25=5u\\\\u=5\;\rm m/s$

Then, the velocity of centre of mass is the sum of the momentum of both bodies divided by the sum of their masses. So,

$v_{cm}=\dfrac{mv'+mu}{m+m} \\\\v_{cm}=\dfrac{5(-1)+(5)\times 5}{5+5} \\\\v_{cm}=\dfrac{20}{10}\\\\v_{cm}=2 \;\rm m/s$

Thus, the velocity of the center of mass of the system at that instant is 2m/s.

Learn more about the center of mass from here:

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answered: leo4687

2m/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.

The move with a common velocity after collision.

Momentum = Mass × velocity.

During the connection of both mass to the string, their total momentum is expressed as:

Mf = (5kg+5kg) × 2

Mf = 20kgm/s

After the spring is released, both objects move remotely with different velocity and have different momentum.

For the block with velocity of 1m/s, its momentum is given as:

M1 = 5×-1

M1 = -5kgm/s(since it's moving to the left)

For the block moving to the right, its momentum is given as:

M2 = 5×u

M2 = 5u

Where u is the velocity of the second block while moving remotely)

Based on the law above;

Mf = M1+M2

20 = -5+5u

20+5 = 5u

u = 25/5

u = 5m/s

The speed of the second block is 5m/s.

To calculate the speed of the center of mass of the system at that instant, we will use the formula:

Centre of mass velocity is the sum of the momentum of both bodies divided by the sum of their masses.

Centre of mass velocity = M1+M2/m1+m2

M1 and M2 are their momentum

m1 and m2 are the masses.

Centre of mass velocity

= -5+5(5)/5+5

= -5+25/10

= 20/10

= 2m/s

Therefore, the speed of the center of mass of the system at that instant is 2m/s

answered: Guest

i would think klcio3

answered: Guest

To use anything transparent

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Two identical blocks with mass 5.0 kg each are connected to the opposite ends of a compressed spring...