---
title: "Direct impact of two spheres"
subject: "Engineering Mechanics"
chapter: 4
key: "semester_3/engineering_mechanics:two-spheres-direct-impact-numerical"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/two-spheres-direct-impact-numerical
site: BCT Engineering
---

# Direct impact of two spheres

Engineering Mechanics · Chapter 4: Kinetics of Particles: Impulse and Momentum

- Asked in: year not given, 8 marks
- Video: [Direct Collision of Two Spheres](https://www.youtube.com/watch?v=ZGdKGESMt3k) — KeysToMaths1
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/two-spheres-direct-impact-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-4

## Question

A sphere A of mass 5 kg moving at 6 m/s overtakes a sphere B of mass 3 kg moving at 2 m/s in the same straight line. The coefficient of restitution is 0.7. Find (a) the velocities of both spheres after impact, (b) the loss of kinetic energy, and (c) the impulse exerted on B.

## Short answer

Applying momentum conservation and restitution e = 0.7 yields v_A = 3.45 m/s, v_B = 6.25 m/s. Kinetic energy loss is 7.65 J. Impulse on B equals its momentum change: 3×(6.25 –2) = 12.75 N·s.

## Answer

**Given:** $m_A = 5$ kg, $u_A = 6$ m/s, $m_B = 3$ kg, $u_B = 2$ m/s, $e = 0.7$ (both to the right).

#### (a) Velocities after impact
Conservation of momentum: total momentum $= 5(6) + 3(2) = 36.0$ kg·m/s.

$$
5v_A + 3v_B = 36 \quad (1)
$$

Restitution: $v_B - v_A = e(u_A - u_B) = 0.7(6 - 2) = 2.8$ (2)

From (2), $v_B = v_A + 2.8$. Put in (1): $5v_A + 3v_A + 8.4 = 36$, so $8v_A = 27.6$.

$$
v_A = 3.45\ \text{m/s} \qquad v_B = 6.25\ \text{m/s}
$$

Both move to the right; B now moves faster than A, so there is no further impact.

#### (b) Loss of kinetic energy

| | Before (J) | After (J) |
|---|---|---|
| A | $\tfrac12(5)(6)^2 = 90$ | $\tfrac12(5)(3.45)^2 = 29.76$ |
| B | $\tfrac12(3)(2)^2 = 6$ | $\tfrac12(3)(6.25)^2 = 58.59$ |
| Total | 96.00 | 88.35 |

$$
\Delta T = 96.00 - 88.35 = 7.65\ \text{J}
$$

Check by formula: $\dfrac{5 \times 3}{2 \times 8}(1 - 0.49)(4)^2 = 0.9375 \times 0.51 \times 16 = 7.65$ J.

#### (c) Impulse on B
Impulse equals change of momentum of B:

$$
I = m_B(v_B - u_B) = 3(6.25 - 2) = 12.75\ \text{N·s}
$$

(The impulse on A is equal and opposite, $5(3.45 - 6) = -12.75$ N·s.)

**Answer: v_A = 3.45 m/s, v_B = 6.25 m/s; energy lost = 7.65 J; impulse on B = 12.75 N·s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the total momentum before impact?
   A. 5(6) - 3(2) = 24
   B. 5(6) + 3(2) = 36 kg m/s
   C. 5 + 3 = 8
   D. (6 + 2)/2 = 4
   Answer: B. Both move to the right, so the momenta add. (Key point: Momentum equation)

2. Which relation gives the second equation?
   A. v_B - v_A = e(u_A - u_B) = 2.8
   B. v_B + v_A = e(u_A + u_B)
   C. v_B - v_A = u_A - u_B
   D. v_A - v_B = 2.8
   Answer: A. Separation speed is e times approach speed (6 - 2 = 4). (Key point: Restitution equation)

3. The velocities after impact are:
   A. v_A = 3.45, v_B = 6.25 m/s
   B. v_A = 4, v_B = 4 m/s
   C. v_A = 2.8, v_B = 6 m/s
   D. v_A = 6.25, v_B = 3.45 m/s
   Answer: A. 8 v_A = 36 - 8.4 = 27.6, so v_A = 3.45 and v_B = 6.25. B must be faster than A after impact. (Key point: Final velocities)

4. The kinetic energy lost is:
   A. 0
   B. 96 J
   C. 7.65 J
   D. 88.35 J
   Answer: C. 96.00 - 88.35 = 7.65 J, matching (m1 m2 / 2(m1+m2))(1 - e^2)(4)^2. (Key point: Energy loss)

## Related in Engineering Mechanics

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- [Oblique impact of ball on a wall](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Aball-oblique-impact-wall) — Engineering Mechanics, chapter 4
- [Kinetic energy and work of rigid body](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Akinetic-energy-rigid-body-work-energy) — Engineering Mechanics, chapter 7

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