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Step-by-Step Solution
Step 1: Write down the known information
• A particle of mass m with initial speed $v_0$ strikes another identical particle (also of mass m) that is initially at rest.
• After the collision, the total kinetic energy increases by 50% of the original kinetic energy.
• We want to find the magnitude of the relative velocity $(v_1 - v_2)$ between the two particles after collision.
Step 2: Use conservation of momentum
Since the collision is collinear and the masses are identical (m each), conserve linear momentum:
$mv_0 = mv_1 + mv_2 \quad \Rightarrow \quad v_0 = v_1 + v_2 \quad (1)$
Step 3: Express the change in kinetic energy
Initially, total kinetic energy is:
$K_i = \tfrac{1}{2} m v_0^2.$
It is given that the final total kinetic energy $K_f$ is 50% greater than $K_i$, which means:
$K_f = \tfrac{3}{2} K_i = \tfrac{3}{2} \times \tfrac{1}{2} m v_0^2 = \tfrac{3}{4} m v_0^2.$
Hence,
$\tfrac{1}{2}m v_1^2 + \tfrac{1}{2}m v_2^2 = \tfrac{3}{4} m v_0^2.$
Dividing both sides by $\tfrac{1}{2}m$:
$v_1^2 + v_2^2 = \tfrac{3}{2}\, v_0^2. \quad (2)$
Step 4: Relate $v_1$ and $v_2$ using momentum equation
From equation (1), we have:
$(v_1 + v_2)^2 = v_0^2.$
Expanding the left side:
$v_1^2 + v_2^2 + 2\,v_1\,v_2 = v_0^2. \quad (3)$
Step 5: Substitute from the kinetic-energy relation
From equation (2): $v_1^2 + v_2^2 = \tfrac{3}{2}\, v_0^2.$
Using this in equation (3):
$\tfrac{3}{2}\, v_0^2 + 2\,v_1\,v_2 = v_0^2.$
So,
$2\,v_1\,v_2 = v_0^2 - \tfrac{3}{2}\, v_0^2 = -\tfrac{1}{2}\, v_0^2.$
Step 6: Find the relative velocity $(v_1 - v_2)$
We use the identity:
$(v_1 - v_2)^2 = (v_1 + v_2)^2 - 4\,v_1\,v_2.$
Substitute the known values:
$(v_1 - v_2)^2 = v_0^2 - 4 \left(\frac{-1}{4}\, v_0^2 \right)
\quad \text{[since } 2\,v_1\,v_2 = -\tfrac{1}{2}\,v_0^2 \text{, hence } v_1\,v_2 = -\tfrac{1}{4}\,v_0^2 \text{]}.$
Hence,
$(v_1 - v_2)^2 = v_0^2 + v_0^2 = 2\, v_0^2.$
So,
$v_1 - v_2 = \sqrt{2}\, v_0.$
Step 7: State the final answer
Therefore, the magnitude of the relative velocity of the two particles after the collision is
$\sqrt{2}\, v_0.$
