Source: Final Exam Practice Problems and Solutions
Tags: linear momentum, conservation of momentum, elastic collision, inelastic collision, kinetic energy transfer, bullet block, 2D collision, support reactions, moment, torque, beam equilibrium, pin support, roller support, ENGR 216, PHYS 216
Conservation of momentum governs collisions: total momentum before equals total momentum after. Whether kinetic energy is conserved too determines if the collision is elastic or inelastic. Support reaction problems extend equilibrium to beams and structures by adding the moment equation, ΣM = 0, to solve for forces at supports like pins and rollers.
Linear momentum
p = mv. A vector quantity, conserved in all collisions when no external forces act.
Conservation of momentum
m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f. Holds for every collision type, in each component direction independently.
Elastic collision
A collision where both momentum and kinetic energy are conserved. No energy is lost to deformation, heat, or sound.
Inelastic collision
A collision where momentum is conserved but kinetic energy is not. Energy is lost to deformation, heat, etc.
Perfectly inelastic collision
The objects stick together after colliding. Maximum kinetic energy loss for a given momentum transfer.
Support reactions
The forces (and sometimes moments) exerted by supports on a structure. Determined by equilibrium: ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0.
Pin support
Provides both a horizontal and vertical reaction force (two unknowns), but no moment.
Roller support
Provides a reaction force perpendicular to the rolling surface only (one unknown).
Moment (torque)
M = r × F. The tendency of a force to cause rotation about a point. In 2D: M = Fd, where d is the perpendicular distance from the point to the line of action.
Given: bullet mass = 5.20 g = 0.00520 kg, v_bullet_initial = 672 m/s, block mass = 700 g = 0.700 kg, v_block_initial = 0, v_bullet_final = 428 m/s.
Apply conservation of momentum:
m_b × v_bi + m_B × v_Bi = m_b × v_bf + m_B × v_Bf
(0.00520)(672) + 0 = (0.00520)(428) + (0.700)(v_Bf)
3.4944 = 2.2256 + 0.700 × v_Bf
v_Bf = 1.2688 / 0.700 = 1.81 m/s
For a 1D elastic collision between mass m (electron) and mass M (hydrogen atom, M = 1840m), with M initially at rest:
Fraction of KE transferred = 4mM / (m + M)²
= 4(1)(1840) / (1 + 1840)² = 7360 / 3,385,281 = 0.00217 = 0.217%
This result is general for any elastic collision with one object initially at rest. When the mass ratio is extreme (m << M), very little energy transfers.
Given: m_A = m_B = 2.0 kg
Before: V_A = 15î + 30ĵ m/s, V_B = −10î + 5ĵ m/s
After: V_A = −5î + 20ĵ m/s, V_B = ?
Conservation of momentum (component by component):
x: 2(15) + 2(−10) = 2(−5) + 2(V_Bx) → 10 = −10 + 2V_Bx → V_Bx = 10
y: 2(30) + 2(5) = 2(20) + 2(V_By) → 70 = 40 + 2V_By → V_By = 15
V_B = 10î + 15ĵ m/s
Kinetic energy change:
KE_before = ½(2)(15² + 30²) + ½(2)(10² + 5²) = 1125 + 125 = 1250 J
KE_after = ½(2)(5² + 20²) + ½(2)(10² + 15²) = 425 + 325 = 750 J
ΔKE = 750 − 1250 = −500 J (500 J lost, so inelastic)
Soccer ball (0.5 kg) held above yoga ball (1.0 kg), dropped from 0.3 m. Both hit the floor, yoga ball transfers all its momentum to the soccer ball.
Velocity at impact: v = √(2gh) = √(2 × 9.81 × 0.3) = 2.426 m/s
Total momentum just before transfer: (m₁ + m₂) × v = 1.5 × 2.426 = 3.639 kg·m/s
All transferred to soccer ball: v_soccer = 3.639 / 0.5 = 7.278 m/s
Height: h = v² / (2g) = 7.278² / (2 × 9.81) = 2.7 m
For a 2D rigid body in static equilibrium, you have three equations:
ΣFₓ = 0
ΣFᵧ = 0
ΣM_about any point = 0
Strategy:
Draw the free body diagram with all applied loads and support reactions
Choose a moment point that eliminates one or more unknowns (typically at a support)
Solve the moment equation first, then use force equations for the remaining unknowns
Applied forces: 500 N horizontal, 350 N vertical downward at C (220 mm from A). Support B provides horizontal force only; support A is a pin (horizontal + vertical).
Taking moments about A to find B, then using force equilibrium:
B = −635î N
A = 135î + 350ĵ N
Beam mass = 120 kg (weight acts at the centre of gravity). External load of 1600 N. Support at A (pin) and B (at 30° angle).
B = −1.71î + 2.96ĵ kN
A = 2.51î − 0.398ĵ kN
Bar in horizontal position, supported by cables AC and BD, with a 300 lb applied load and unknown force P.
T_A = 346 lbf, T_B = 462 lbf, P = −429ĵ lbf
Conservation of momentum: Σmᵢvᵢ (before) = Σmᵢvᵢ (after)
Kinetic energy: KE = ½mv² (translational)
Elastic collision energy transfer fraction (1D, one at rest): 4mM / (m + M)²
Free fall velocity: v = √(2gh)
Maximum height from velocity: h = v² / (2g)
Equilibrium equations (2D): ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
Moment about a point: M = F × d (perpendicular distance)
⚠️ Momentum is conserved in all collisions. Kinetic energy is conserved only in elastic collisions. You must check.
⚠️ In 2D collisions, apply conservation of momentum separately in x and y. You get two equations.
⚠️ When computing KE in 2D, use v² = vₓ² + vᵧ². Do not forget to include both components.
⚠️ For the elastic collision energy transfer formula, it applies only when one object starts at rest.
⚠️ In support reaction problems, always take moments about a point where an unknown force acts. This eliminates that unknown from the moment equation, making the algebra simpler.
⚠️ Watch the sign conventions. Pick a consistent positive direction for forces and a consistent rotation sense for moments, and stick with it.
Q: A 5.20 g bullet at 672 m/s passes through a 700 g block at rest and exits at 428 m/s. What is the block's speed?
A: Conservation of momentum: (0.0052)(672) = (0.0052)(428) + (0.700)(v). Solving gives v = 1.81 m/s.
Q: In an elastic collision where m << M (M at rest), approximately what fraction of KE is transferred?
A: Approximately 4m/M. For an electron hitting a hydrogen atom (M = 1840m), that is about 4/1840 ≈ 0.217%.
Q: Two 2.0 kg objects collide. Total KE drops from 1250 J to 750 J. Is this elastic or inelastic?
A: Inelastic. 500 J of kinetic energy was lost.
Q: Why do you take moments about a support point in equilibrium problems?
A: The unknown reaction force at that point has zero moment arm, so it drops out of the equation, leaving fewer unknowns to solve.
Q: A 0.5 kg ball receives all the momentum from a 1.5 kg system falling from 0.3 m. How high does it bounce?
A: Impact velocity = √(2 × 9.81 × 0.3) = 2.43 m/s. Total momentum = 1.5 × 2.43 = 3.64 kg·m/s. Soccer ball velocity = 3.64/0.5 = 7.28 m/s. Height = 7.28²/(2 × 9.81) = 2.7 m.
linear momentum, conservation of momentum, elastic collision, inelastic collision, perfectly inelastic, kinetic energy transfer, bullet block problem, 2D collision, vector momentum, stacked ball drop, support reactions, beam equilibrium, pin support, roller support, free body diagram, moment, torque, perpendicular distance, cable tension, static equilibrium, ENGR 216 Module 8, ENGR 216 Module 9