Physics Q88 – AP-EAMCET 2026

Home Physics Work, Energy & PowerQuestion #640

A ball of mass M moving with a velocity of

4 {m s}^{- 1}

collides with another ball of mass

2 M

moving in the same direction. If the coefficient of restitution between the two balls is

0.35

and the velocity of the heavier ball after collision is

2.9 {m s}^{- 1}

, then the initial velocity of the heavier ball is

Clue 1 of 3: Conservation Laws and Restitution

For a collision between two bodies, the total linear momentum of the system is conserved, and the relative velocities before and after the collision are related by the coefficient of restitution.

🧠 Full Solution Path

Step 1: Define Variables and Given Data✦ Active

Let the first ball (lighter) have mass $m_{1} = M$ and initial velocity $u_{1} = 4 {m s}^{- 1}$ . Let its final velocity be $v_{1}$ .

Let the second ball (heavier) have mass $m_{2} = 2 M$ and initial velocity $u_{2}$ . Let its final velocity be $v_{2} = 2.9 {m s}^{- 1}$ .

The coefficient of restitution is $e = 0.35$ . Both balls move in the same direction initially.

We need to find the initial velocity of the heavier ball, $u_{2}$ .

Step 2: Formulate Momentum Conservation Equation○ Expand

According to the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision:

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Substituting the given values into the momentum equation:

M (4) + 2 M (u_{2}) = M (v_{1}) + 2 M (2.9)

Dividing by $M$ and simplifying, we get an expression for $v_{1}$ :

4 + 2 u_{2} = v_{1} + 5.8

v_{1} = 2 u_{2} - 1.8 (Equation 1)

💡 Teacher's Secret Hint

Remember that momentum is a vector quantity. Since both balls are moving in the same direction, we can assign a positive sign to their velocities.

Step 3: Formulate Coefficient of Restitution Equation○ Expand

The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:

e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}

Substituting the known values:

0.35 = \frac{2.9 - v_{1}}{4 - u_{2}}

Rearranging this equation to express $v_{1}$ :

0.35 (4 - u_{2}) = 2.9 - v_{1}

1.4 - 0.35 u_{2} = 2.9 - v_{1}

v_{1} = 2.9 - 1.4 + 0.35 u_{2}

v_{1} = 1.5 + 0.35 u_{2} (Equation 2)

💡 Teacher's Secret Hint

Ensure the correct order for relative velocities ( $v_{2} - v_{1}$ and $u_{1} - u_{2}$ ) to maintain the positive value of $e$ as given.

Step 4: Solve for the Unknown Initial Velocity○ Expand

Now, we equate the two expressions for $v_{1}$ (Equation 1 and Equation 2):

2 u_{2} - 1.8 = 1.5 + 0.35 u_{2}

Group terms involving $u_{2}$ and constant terms:

2 u_{2} - 0.35 u_{2} = 1.5 + 1.8

1.65 u_{2} = 3.3

Solve for $u_{2}$ :

u_{2} = \frac{3.3}{1.65}

u_{2} = 2 {m s}^{- 1}

The initial velocity of the heavier ball is $2 {m s}^{- 1}$ .

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Physics Question 88 – AP-EAMCET 2026

Clue 1 of 3: Conservation Laws and Restitution

🧠 Full Solution Path