Based on the analysis of velocities and the conservation of momentum, block 2 is estimated to be five times heavier than block 1. This aligns with the observed behavior in the collision dynamics. Here option D is correct.
Block 2 has the greater mass. This is indicated by the fact that the velocity of block 2 is closer to the initial velocity of block 1 than it is to the initial velocity of block 2. This means that block 2 was able to slow down block 1 more than block 1 was able to speed up block 2.
Conservation of momentum. In a collision between two objects on a frictionless surface, the total momentum of the system is conserved.
This means that the sum of the products of mass and velocity of the objects before the collision is equal to the sum of the products of mass and velocity of the objects after the collision. Mathematically, this can be expressed as:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁ and v₂ are the initial velocities of block 1 and block 2, respectively, and v₁' and v₂' are the final velocities of block 1 and block 2, respectively.
Solving for mass ratio. Using the conservation of momentum equation, we can solve for the ratio of the masses of the two blocks. We can assume that block 1 is moving to the right and block 2 is moving to the left before the collision, and vice versa after the collision. Therefore, we can write:
m₁v₁ - m₂v₂ = -m₁v₁' + m₂v₂'
Dividing both sides by v₁, we get:
m₁ - m₂(v₂/v₁) = -m₁(v₁'/v₁) + m₂(v₂'/v₁)
Rearranging the terms, we get:
m₁(1 + v₁'/v₁) = m₂(v₂/v₁ + v₂'/v₁)
Dividing both sides by m₁, we get:
1 + v₁'/v₁ = m₂/m₁(v₂/v₁ + v₂'/v₁)
Finally, solving for m₂/m₁, we get:
(m₂/m₁) = (1 + v₁'/v₁) / (v₂/v₁ + v₂'/v₁)
Using the graph values. From the graph, we can estimate the values of the velocities of the blocks before and after the collision. For example, we can use the following values:
v₁ ≈ 4 m/s
v₂ ≈ -2 m/s
v₁' ≈ 1 m/s
v₂' ≈ 3 m/s
Plugging these values into the formula for the mass ratio, we get:
(m₂/m₁) ≈ (1 + 1/4) / (-2/4 + 3/4) ≈ (5/4) * (4/1) = 5
This means that block 2 is five times heavier than block 1. This is consistent with the observation that block 2 changed its velocity less than block 1 after the collision. Here option D is correct.