Question

Three blocks are initially placed as shown in the figure. Block A has mass m and initial velocity v to the right. Block B with mass m and block C with mass 4m are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block A is

• A. 0.6v to the left
• B. 1.4v to the left
• C. v to the left
• D. 0.4v to the right

Answer 1

Answer: (A)

0.6v to the left

Answer 2

The problem involves a series of elastic collisions between three blocks A, B, and C. We need to find the final velocity of block A.

Given:

• Mass of block A ($m_A$) = m
• Mass of block B ($m_B$) = m
• Mass of block C ($m_C$) = 4m
• Initial velocity of block A ($u_A$) = v (to the right, let's take right as positive direction)
• Initial velocity of block B ($u_B$) = 0
• Initial velocity of block C ($u_C$) = 0
• All collisions are elastic, and friction is neglected.

For an elastic collision between two bodies with masses $m_1$ and $m_2$ and initial velocities $u_1$ and $u_2$, the final velocities $v_1$ and $v_2$ are given by: $v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}$ $v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}$

Step 1: Collision between Block A and Block B

Block A (mass m, velocity v) collides with Block B (mass m, velocity 0). Since the masses are equal ($m_A = m_B = m$) and the collision is elastic, the blocks will exchange velocities.

• Final velocity of A ($v_A'$) = initial velocity of B = 0
• Final velocity of B ($v_B'$) = initial velocity of A = v

After this collision:

• Block A is at rest ($v_A' = 0$).
• Block B moves to the right with velocity $v$ ($v_B' = v$).

Step 2: Collision between Block B and Block C

Now, Block B (mass $m_B = m$, velocity $u_B'' = v$) collides with Block C (mass $m_C = 4m$, velocity $u_C'' = 0$). Using the elastic collision formulas:

• Final velocity of B ($v_B''$): $v_B'' = \frac{(m_B - m_C)u_B'' + 2m_C u_C''}{m_B + m_C} = \frac{(m - 4m)v + 2(4m)(0)}{m + 4m} = \frac{-3mv}{5m} = -0.6v$
• Final velocity of C ($v_C''$): $v_C'' = \frac{(m_C - m_B)u_C'' + 2m_B u_B''}{m_B + m_C} = \frac{(4m - m)(0) + 2m v}{m + 4m} = \frac{2mv}{5m} = 0.4v$

After this collision:

• Block B moves to the left with velocity $0.6v$ ($v_B'' = -0.6v$).
• Block C moves to the right with velocity $0.4v$ ($v_C'' = 0.4v$).
• Block A is still at rest ($v_A' = 0$).

Step 3: Collision between Block B and Block A (again)

Now, Block B (mass $m_B = m$, velocity $u_B''' = -0.6v$) collides with Block A (mass $m_A = m$, velocity $u_A''' = 0$). Since the masses are equal ($m_A = m_B = m$) and the collision is elastic, the blocks will exchange velocities.

• Final velocity of B ($v_B'''$) = initial velocity of A = 0
• Final velocity of A ($v_A'''$) = initial velocity of B = $-0.6v$

After this collision:

• Block A moves to the left with velocity $0.6v$ ($v_A''' = -0.6v$).
• Block B is at rest ($v_B''' = 0$).
• Block C continues to move to the right with velocity $0.4v$.

At this point, Block A is moving left, Block B is at rest, and Block C is moving right. No further collisions will occur between A, B, and C. Therefore, the final velocity of block A is $0.6v$ to the left.