---
title: "Impulse-momentum principle; blocks on incline"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:impulse-momentum-blocks-a-b-incline-pulley"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/impulse-momentum-blocks-a-b-incline-pulley
site: BCT Engineering
---

# Impulse-momentum principle; blocks on incline

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2070 Chaitra, 2+8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/impulse-momentum-blocks-a-b-incline-pulley
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

What do you mean by impulse momentum principle? Two blocks A and B having respective weights 500 N and 1000 N start from rest. The pulley is frictionless and also practically mass less. The kinetic coefficient of friction between the block A and the inclined surface is 0.35. Determine the acceleration of each block and tension in the cord. [Figure: block A (500 N) on a 15° inclined surface connected by a cord over a pulley to hanging block B (1000 N).]

## Short answer

Linear impulse ∫F dt equals change in linear momentum m v₂ − m v₁. For the blocks: on A, tension minus friction and weight component along incline = mass×a; on B, weight minus tension = mass×a. Solving gives acceleration 4.59 m/s² for both and tension 532.3 N.

## Answer

#### Impulse–momentum principle
The linear impulse of the resultant force acting on a particle during a time interval is equal to the change in its linear momentum:

∫ F dt = m v₂ − m v₁ (or m v₁ + ∫ F dt = m v₂)

Linear impulse has units N·s and equals the area under the force–time graph. If the resultant force is zero the momentum is conserved. For a constant force, F t = m(v₂ − v₁), which is Newton's second law in integrated form.

#### Blocks A and B
A (500 N) lies on a 15° incline and is joined by a cord over a frictionless, massless pulley to B (1000 N) hanging vertically. B is heavier, so B descends and A is pulled up the incline. μ_k = 0.35. Let a be the common acceleration and T the cord tension.

- Normal reaction on A: N = 500 cos 15° = 482.96 N
- Friction on A (down the incline): F = 0.35 × 482.96 = 169.04 N
- Component of A's weight along the incline: 500 sin 15° = 129.41 N

```text
   A \ (500 N)        15° incline, A moves up
      \ ___ o pulley
              |
              B (1000 N) moves down
```

**Block B:** 1000 − T = (1000/9.81) a
**Block A:** T − 129.41 − 169.04 = (500/9.81) a

Adding: 1000 − 298.45 = (1500/9.81) a
701.55 = 152.91 a

**a = 4.59 m/s²** for both blocks (A up the slope, B down).

Tension: T = 1000 − (1000/9.81)(4.588) = **532.3 N**

**Answer: a_A = a_B = 4.59 m/s²; T = 532.3 N.**

## Check yourself

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

1. What does the impulse-momentum principle say?
   A. Linear impulse of the resultant force equals the change in linear momentum
   B. Impulse equals change in kinetic energy
   C. Momentum is always conserved
   D. Impulse is force times distance
   Answer: A. Integral of F dt = m v2 - m v1. Momentum is conserved only if the resultant force is zero. (Key point: Impulse-momentum principle)

2. How is the friction on A computed?
   A. 0.35 x 500 = 175 N
   B. 0.35 x 500 sin 15 = 45.3 N
   C. 0.35 x 500 cos 15 = 169.04 N
   D. 0.35 x 1000 = 350 N
   Answer: C. The normal reaction on an incline is W cos 15 = 482.96 N, so F = 0.35 x 482.96 = 169.04 N, acting down the slope. (Key point: Friction on the incline)

3. Why does A move up the incline, and what does the equation for A include?
   A. B is heavier; T - 500 sin 15 - friction = (500/g)a
   B. A is lighter so it slides down; T + friction = m a
   C. Friction pulls A up the slope
   D. Tension is equal to 1000 N
   Answer: A. B (1000 N) outweighs the resistance on A, so A moves up. Both the weight component 129.41 N and friction 169.04 N oppose it. (Key point: Direction and equations)

4. Adding the equations for A and B gives which result?
   A. a = 1000 x 9.81 / 500
   B. a = 701.55 / (1500/9.81) = 4.59 m/s^2
   C. a = 1000 / 298.45
   D. a = 298.45 x 9.81 / 1500
   Answer: B. 1000 - 129.41 - 169.04 = (1500/g)a, so a = 4.59 m/s^2 for both blocks, and T = 532.3 N. (Key point: Eliminating T)

## Related in Applied Mechanics

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- [Block sliding on incline compressing spring](https://www.bctengineer.com/question.md?key=semester_3%2Fengineering_mechanics%3Ablock-incline-spring-compression) — Engineering Mechanics, chapter 3
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