Module 2: Dynamics

The resultant of the two forces is $\sqrt{6^2 + 8^2} = 10\text{ N}$, directed $\tan^{-1}(8/6) \approx 53^\circ$ north of east. The equilibrant must be equal in size but exactly opposite, so it is $10\text{ N}$ at $53^\circ$ south of west. (An equilibrant always opposes the resultant.)

First combine the perpendicular forces: $F_{net} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ N}$. Then $a = \dfrac{F_{net}}{m} = \dfrac{13}{3.0} \approx 4.3\text{ m/s}^2$. (Adding the forces as $17\text{ N}$ ignores that they are at right angles.)

Along the slope: $a = g(\sin\theta - \mu\cos\theta) = 9.8(\sin 30^\circ - 0.20\cos 30^\circ) = 9.8(0.5 - 0.173) = 3.2\text{ m/s}^2$. The mass cancels. (Ignoring friction gives $4.9\text{ m/s}^2$; friction removes about $1.7\text{ m/s}^2$.)

At the point of slipping, the friction is at its maximum and balances the component of weight along the slope: $\mu mg\cos\theta = mg\sin\theta$. So $\mu = \tan\theta = \tan 30^\circ \approx 0.58$. The mass and $g$ cancel, leaving only the angle.

Vertically, the two ropes together support the weight: $2T\sin 30^\circ = mg$. So $2T(0.5) = 10 \times 9.8$, i.e. $T = 98\text{ N}$. At this shallow angle each rope carries the full weight, not half of it, a result that surprises many.

An action-reaction pair acts on two different objects and is the same type of force. The Earth pulls the book down (gravity), so the reaction is the book pulling the Earth up (also gravity). The table's normal force balances the weight but is a separate contact interaction, not the third-law partner.

The hanging weight drives the system; friction on the block opposes it. $F_{net} = m_2 g - \mu m_1 g = 2.0(9.8) - 0.25(3.0)(9.8) = 19.6 - 7.35 = 12.25\text{ N}$. Then $a = \dfrac{12.25}{m_1 + m_2} = \dfrac{12.25}{5.0} = 2.45\text{ m/s}^2$. (Ignoring friction gives $3.92\text{ m/s}^2$.)

The lift moves down but slows, so its acceleration points upward: $a = +1.5\text{ m/s}^2$. Then $N = m(g + a) = 60(9.8 + 1.5) = 60 \times 11.3 = 678\text{ N}$. The person feels heavier, not lighter, even though the lift is going down (the direction of acceleration, not velocity, is what matters).

Newton's second law upward: $T - mg = ma$. At the breaking point $T = 25\text{ N}$, so $25 - 2.0(9.8) = 2.0a$, giving $a = \dfrac{25 - 19.6}{2.0} = 2.7\text{ m/s}^2$. (You must subtract the weight; using $25/2 = 12.5$ forgets gravity.)

First find the system acceleration: $a = \dfrac{F}{m_1 + m_2} = \dfrac{12}{6.0} = 2.0\text{ m/s}^2$. The contact force is the only horizontal force on the front $2.0\text{ kg}$ block, so $F_c = m_2 a = 2.0 \times 2.0 = 4.0\text{ N}$. (It is not $12\text{ N}$: only part of the push is needed to accelerate the smaller block.)

Energy: $mgh - W_{friction} = \tfrac{1}{2}mv^2$. So $2.0(9.8)(3.0) - 20 = 58.8 - 20 = 38.8\text{ J}$ $= \tfrac{1}{2}(2.0)v^2$, giving $v^2 = 38.8$ and $v \approx 6.2\text{ m/s}$. (Ignoring friction gives $7.7\text{ m/s}$.)

The braking force does work equal to the kinetic energy: $Fd = \tfrac{1}{2}mv^2$. So $d = \dfrac{\tfrac{1}{2}(1200)(20)^2}{6000} = \dfrac{240000}{6000} = 40\text{ m}$. (Using energy avoids first finding the deceleration.)

The braking force does work equal to the kinetic energy: $Fd = \tfrac{1}{2}mv^2$, so $d \propto v^2$. Doubling $v$ multiplies $v^2$ (and therefore $d$) by 4. This is why a small increase in speed sharply increases stopping distance.

The driving force must overcome resistance and also produce the acceleration: $F = ma + F_{resist} = 1000(2.0) + 500 = 2500\text{ N}$. Then $P = Fv = 2500 \times 15 = 37500\text{ W}$ = $37.5\text{ kW}$. (Using only $ma$ or only the resistance misses part of the force.)

Energy is conserved on the smooth swing: $\tfrac{1}{2}mv^2 = mgh$, so $v = \sqrt{2gh} = \sqrt{2(9.8)(0.80)} = \sqrt{15.68} \approx 4.0\text{ m/s}$. The mass and the string length do not affect the answer, only the height dropped.

Useful (output) power is the rate of gaining gravitational PE: $P_{out} = mgv = 20(9.8)(2.0) = 392\text{ W}$. Efficiency $= \dfrac{P_{out}}{P_{in}} = \dfrac{392}{500} = 0.78$, i.e. 78%. The missing $108\text{ W}$ is lost as heat and sound.

Friction does work equal to the kinetic energy removed: $f \times d = \tfrac{1}{2}mv^2$. So $f = \dfrac{\tfrac{1}{2}(3.0)(6.0)^2}{4.0} = \dfrac{54}{4.0} = 13.5\text{ N}$. (Energy avoids needing the acceleration first.)

Two stages. Momentum (collision): $v = \dfrac{0.020 \times 300}{2.02} \approx 2.97\text{ m/s}$. Then energy (swing up): $h = \dfrac{v^2}{2g} = \dfrac{2.97^2}{19.6} \approx 0.45\text{ m}$. You cannot use the bullet's kinetic energy directly, because most of it is lost as heat when it embeds.

Impulse is the change in momentum, and velocity reverses direction. Taking the rebound direction as positive: $J = m(v - u) = 0.15(15 - (-20)) = 0.15 \times 35 = 5.25\text{ N}$·s. (Using $20 - 15 = 5$ treats the speeds as if in the same direction, giving the wrong $0.75\text{ N}$·s.)

Momentum is conserved separately in each direction. East: $2.0 \times 3.0 = 6.0\text{ kg}$·m/s; north: $3.0 \times 2.0 = 6.0\text{ kg}$·m/s. Total momentum $= \sqrt{6^2 + 6^2} = 8.49\text{ kg}$·m/s, shared by $5.0\text{ kg}$, so $v = \dfrac{8.49}{5.0} \approx 1.7\text{ m/s}$. (Momentum adds as a vector, not as a simple sum.)

For equal masses in a head-on elastic collision, the two balls exchange velocities. Momentum and kinetic energy are both conserved, and the only solution satisfying both is that the first ball stops and the second leaves at the original speed. This is what you see in Newton's cradle.

Impulse is the area of the trapezium: $J = \tfrac{1}{2}(b_1 + b_2)h = \tfrac{1}{2}(0.7 + 0.3)(10) = 5.0\text{ N}$·s (parallel sides $0.7\text{ s}$ at the base and $0.3\text{ s}$ across the flat top). Then $v = \dfrac{J}{m} = \dfrac{5.0}{2.5} = 2.0\text{ m/s}$. (Using the peak force of $10\text{ N}$ directly is the trap.)

First momentum gives the common speed: $v = \dfrac{2.0 \times 5.0}{5.0} = 2.0\text{ m/s}$. Kinetic energy before $= \tfrac{1}{2}(2.0)(5.0)^2 = 25\text{ J}$; after $= \tfrac{1}{2}(5.0)(2.0)^2 = 10\text{ J}$. Energy lost $= 15\text{ J}$ (converted to heat and sound).

Momentum first: $v = \dfrac{2.0 \times 6.0}{6.0} = 2.0\text{ m/s}$. Then friction decelerates them at $a = \mu g = 0.30 \times 9.8 = 2.94\text{ m/s}^2$, so $d = \dfrac{v^2}{2a} = \dfrac{2.0^2}{5.88} \approx 0.68\text{ m}$. (Using the $6.0\text{ m/s}$ in the second stage forgets the collision slows them first.)

The occupant's change in momentum (from crash speed to zero) is fixed, so the impulse $Ft$ is fixed. By extending the stopping time $t$, the airbag makes the average force $F = \dfrac{\Delta p}{t}$ much smaller. Crumple zones and seatbelts work on the same principle.