Apply Forces & Momentum - IB Physics

1. A 20 kg crate is pushed 5.0 m up a rough incline at a constant velocity by a force parallel to the incline. The incline is at 30° to the horizontal and the coefficient of kinetic friction is 0.20. What is the work done by the applied force? (Use $g = 10 \text{ m s}^{-2}$)
   Answer: 670 J Since velocity is constant, the net force is zero. The applied force $F_{app}$ must balance the force of friction $F_f$ and the component of gravity parallel to the incline, $mg\sin\theta$. $F_{app} = F_f + mg\sin\theta$. The normal force is $F_N = mg\cos\theta = 20 \times 10 \times \cos(30^{\circ}) = 173.2\text{ N}$. The friction force is $F_f = \mu_k F_N = 0.20 \times 173.2 = 34.64\text{ N}$. The gravity component is $mg\sin\theta = 20 \times 10 \times \sin(30^{\circ}) = 100\text{ N}$. So, $F_{app} = 34.64 + 100 = 134.64\text{ N}$. The work done is $W = F_{app} \times d = 134.64 \times 5.0 = 673.2\text{ J}$, which is approximately 670 J.
2. A 0.5 kg ball moving at 8.0 m/s collides with a wall at an angle of 60° to the normal. The ball rebounds at 40° to the normal with speed 6.0 m/s. What is the magnitude of the impulse delivered by the wall to the ball?
   Answer: 5.9 kg⋅m/s at an angle to the wall normal Break into components parallel and perpendicular to wall. Initial: $p_{i,\parallel} = 0.5 \times 8.0 \times \sin(60°) = 3.46$ kg⋅m/s, $p_{i,\perp} = 0.5 \times 8.0 \times \cos(60°) = 2.0$ kg⋅m/s (toward wall). Final: $p_{f,\parallel} = 0.5 \times 6.0 \times \sin(40°) = 1.93$ kg⋅m/s (same direction), $p_{f,\perp} = -0.5 \times 6.0 \times \cos(40°) = -2.30$ kg⋅m/s (away from wall). Impulse components: $J_{\parallel} = 1.93 - 3.46 = -1.53$ kg⋅m/s, $J_{\perp} = -2.30 - 2.0 = -4.30$ kg⋅m/s. Magnitude: $|J| = \sqrt{(-1.53)^2 + (-4.30)^2} = 5.9$ kg⋅m/s. Choice A uses only parallel component. Choice B uses only perpendicular component incorrectly. Choice D uses incorrect angle calculations.
3. A block of mass $m$ rests on the floor of a lift that is accelerating upwards with acceleration $a$. A horizontal force $F$ is applied to the block, but the block does not move. The coefficient of static friction between the block and the floor is $\mu_s$. What is the maximum possible value of $F$ for which the block will not move?
   Answer: $\mu_s m (g+a)$ The net vertical force on the block is $F_N - mg = ma$, where $F_N$ is the normal force. Therefore, $F_N = m(g+a)$. The maximum static friction force is $F_{f,max} = \mu_s F_N$. For the block not to move, the applied horizontal force $F$ must be less than or equal to the maximum static friction force. The maximum possible value of $F$ is thus $F_{max} = \mu_s F_N = \mu_s m(g+a)$.
4. A sphere of mass $m$ moving at speed $v$ makes a head-on elastic collision with a sphere of mass (2m) that is initially at rest. What is the speed of the mass (2m) after the collision?
   Answer: $2v/3$ For a head-on elastic collision, both momentum and kinetic energy are conserved. Using the conservation laws, the final velocity $v_2'$ of the target mass $m_2 = 2m$ can be found. A useful formula derived from these laws is $v_2' = \frac{2m_1}{m_1+m_2}v_1$, where $v_1=v$ is the initial velocity of the projectile. Substituting the masses: $v_2' = \frac{2m}{m+2m}v = \frac{2m}{3m}v = \frac{2}{3}v$.
5. A child and a skateboard have a combined mass of 49 kg. They are stationary on a frictionless horizontal surface. The child throws a 1.0 kg ball horizontally with a speed of 50 m s⁻¹ relative to the skateboard. What is the magnitude of the final velocity of the skateboard?
   Answer: 1.0 m s⁻¹ Let $M$ be the mass of the child and skateboard (49 kg) and $m$ be the mass of the ball (1.0 kg). Let $v_s$ be the final velocity of the skateboard and $v_b$ be the final velocity of the ball, both relative to the ground. The velocity of the ball relative to the skateboard is $v_{rel} = v_b - v_s = 50 \text{ m s}^{-1}$. So, $v_b = 50 + v_s$. Total momentum is conserved, and the initial momentum is zero. $Mv_s + mv_b = 0$. Substituting for $v_b$: $Mv_s + m(50 + v_s) = 0$. $(M+m)v_s + 50m = 0$. $v_s = -\frac{50m}{M+m} = -\frac{50 \times 1.0}{49+1.0} = -\frac{50}{50} = -1.0 \text{ m s}^{-1}$. The magnitude is 1.0 m s⁻¹.
6. Three blocks with masses $m_1 = 1.0$ kg, $m_2 = 2.0$ kg, and $m_3 = 3.0$ kg are connected by light, inextensible strings. The blocks are pulled along a frictionless horizontal surface by a horizontal force $F = 12$ N applied to $m_3$. What is the tension in the string connecting $m_1$ and $m_2$?
   Answer: 2.0 N First, find the acceleration of the entire system. The total mass is $M = m_1 + m_2 + m_3 = 1.0 + 2.0 + 3.0 = 6.0 \text{ kg}$. The net force on the system is $F = 12 \text{ N}$. The acceleration is $a = F/M = 12 \text{ N} / 6.0 \text{ kg} = 2.0 \text{ m s}^{-2}$. The tension in the string between $m_1$ and $m_2$, let's call it $T_1$, is the force that accelerates $m_1$. Applying Newton's second law to $m_1$ alone: $T_1 = m_1 a = 1.0 \text{ kg} \times 2.0 \text{ m s}^{-2} = 2.0 \text{ N}$.
7. A mass $m$ is attached to a vertical spring with spring constant $k$. The mass is at rest at its equilibrium position. It is then displaced downwards by a distance $A$ and released. What is the magnitude of the net force on the mass when it is at a distance $A/2$ above the equilibrium position?
   Answer: $kA/2$ At the equilibrium position, the upward spring force balances the downward force of gravity, $kx_0 = mg$, where $x_0$ is the equilibrium extension. When the mass is a distance $y = A/2$ above equilibrium, the spring extension is $x_0 - y$. The upward spring force is $F_{spring} = k(x_0 - y) = kx_0 - ky = mg - ky$. The net force is $F_{net} = F_{spring} - mg = (mg - ky) - mg = -ky$. The magnitude of this restoring force is $ky$. Substituting $y=A/2$, the magnitude of the net force is $k(A/2)$.
8. An object of mass 2.0 kg is initially at rest. A time-varying horizontal force acts on it. The force starts at 0 N, increases linearly to 10 N over 2.0 s, and then decreases linearly to 0 N over the next 2.0 s. What is the final speed of the object, assuming no friction?
   Answer: 10 m s⁻¹ The change in momentum of the object is equal to the impulse, which is the area under the force-time graph. The graph described is a triangle with a base of 4.0 s (2.0 s + 2.0 s) and a height of 10 N. The area is $\text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \text{ s} \times 10 \text{ N} = 20 \text{ Ns}$. Since impulse equals change in momentum ($J = \Delta p = m\Delta v$) and the object starts from rest, $20 \text{ Ns} = (2.0 \text{ kg}) \times v_f$. Solving for the final speed gives $v_f = 20 / 2.0 = 10 \text{ m s}^{-1}$.
9. A large truck of mass $M$ collides with a small car of mass $m$ (where $M > m$). During the collision, which of the following statements is always correct?
   Answer: The change in momentum of the truck is equal in magnitude to the change in momentum of the car. According to Newton's third law, the force the truck exerts on the car is equal in magnitude and opposite in direction to the force the car exerts on the truck. Since the forces are equal and act for the same time interval, the impulses ($F\Delta t$) on both vehicles are equal in magnitude. Since impulse is equal to the change in momentum, the magnitudes of the changes in momentum are equal.
10. A block of mass $m$ is on a horizontal surface with a coefficient of static friction $\mu_s$. A pulling force $F$ is applied at an angle $\theta$ above the horizontal. What is the minimum magnitude of $F$ that will cause the block to start moving?
    Answer: $\frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$ To start moving, the horizontal component of the applied force must overcome static friction: $F\cos\theta \ge F_f$. The normal force $F_N$ is reduced by the upward component of $F$, so $F_N + F\sin\theta = mg$, which means $F_N = mg - F\sin\theta$. The maximum static friction is $F_{f,max} = \mu_s F_N = \mu_s(mg - F\sin\theta)$. Setting the forces equal for the minimum condition: $F\cos\theta = \mu_s(mg - F\sin\theta)$. Rearranging to solve for F: $F\cos\theta = \mu_s mg - \mu_s F\sin\theta$, so $F(\cos\theta + \mu_s \sin\theta) = \mu_s mg$. Thus, $F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta}$.
11. A block of mass $m$ moving with speed $v$ makes a head-on perfectly inelastic collision with a stationary block of mass (3m). The two blocks stick together. What fraction of the initial kinetic energy is converted to other forms of energy?
    Answer: 3/4 Initial momentum is $p_i = mv$. Initial kinetic energy is $KE_i = \frac{1}{2}mv^2$. After the collision, the combined mass is (4m). By conservation of momentum, $mv = (4m)v_f$, so the final velocity is $v_f = v/4$. The final kinetic energy is $KE_f = \frac{1}{2}(4m)v_f^2 = \frac{1}{2}(4m)(v/4)^2 = \frac{1}{2}(4m)(v^2/16) = \frac{1}{8}mv^2$. The energy lost is $\Delta KE = KE_i - KE_f = \frac{1}{2}mv^2 - \frac{1}{8}mv^2 = \frac{3}{8}mv^2$. The fraction of energy lost is $\frac{\Delta KE}{KE_i} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4}$.
12. An object of mass 5.0 kg is initially at rest. It is subjected to two perpendicular forces, $F_1 = 30$ N and $F_2 = 40$ N. What is the magnitude of the object's acceleration?
    Answer: 10 m s⁻² The net force on the object is the vector sum of the two perpendicular forces. The magnitude of the net force can be found using the Pythagorean theorem: $F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ N}$. According to Newton's second law, $F_{net} = ma$. Therefore, the acceleration is $a = F_{net} / m = 50 \text{ N} / 5.0 \text{ kg} = 10 \text{ m s}^{-2}$.
13. A block of mass 3.0 kg rests on a rough horizontal table (coefficient of kinetic friction 0.20). It is connected by a light string over a frictionless pulley to a hanging block of mass 1.0 kg. The system is released from rest. What is the acceleration of the system? (Use $g = 10 \text{ m s}^{-2}$)
    Answer: 1.0 m s⁻² The driving force is the weight of the hanging block: $F_{drive} = m_{hang}g = 1.0 \times 10 = 10 \text{ N}$. The opposing friction force is $F_f = \mu_k F_N = \mu_k m_{table}g = 0.20 \times 3.0 \times 10 = 6.0 \text{ N}$. The net force on the system is $F_{net} = F_{drive} - F_f = 10 - 6.0 = 4.0 \text{ N}$. This net force accelerates the total mass of the system, $M_{total} = m_{hang} + m_{table} = 1.0 + 3.0 = 4.0 \text{ kg}$. Using Newton's second law, $a = F_{net} / M_{total} = 4.0 / 4.0 = 1.0 \text{ m s}^{-2}$.
14. Two blocks connected by a light rope slide down a frictionless incline. Block A (3.0 kg) is higher on the 30° incline, and block B (2.0 kg) is lower. A horizontal force of 15 N is applied to block A parallel to the incline surface, up the incline. What is the tension in the rope connecting the blocks?
    Answer: 6.0 N because tension depends on the net acceleration of the system First find system acceleration. Total mass = 5.0 kg. Forces down incline: $(m_A + m_B)g\sin(30°) = 5.0 \times 9.8 \times 0.5 = 24.5$ N. Force up incline: 15 N. Net force down incline: $24.5 - 15 = 9.5$ N. Acceleration down incline: $a = \frac{9.5}{5.0} = 1.9$ m/s². For block B alone: $m_Bg\sin(30°) - T = m_Ba$, so $T = 2.0 \times 9.8 \times 0.5 - 2.0 \times 1.9 = 9.8 - 3.8 = 6.0$ N. Choice A incorrectly averages forces. Choice C incorrectly applies the external force to tension calculation. Choice D only considers mass ratio without proper force analysis.
15. An object with a volume of $5.0 \times 10^{-4} \text{ m}^3$ and a density of 800 kg m⁻³ is fully submerged in water (density 1000 kg m⁻³). It is held stationary by a vertical string attached to the bottom of the tank. What is the tension in the string? (Use $g = 10 \text{ m s}^{-2}$)
    Answer: 1.0 N The forces acting on the object are its weight (downwards), the buoyant force (upwards), and the tension in the string (downwards). For equilibrium, the upward force equals the sum of the downward forces: $F_b = W + T$. The weight is $W = m_{obj}g = (\rho_{obj}V)g = (800 \times 5.0 \times 10^{-4}) \times 10 = 4.0 \text{ N}$. The buoyant force is $F_b = \rho_{water}Vg = (1000 \times 5.0 \times 10^{-4}) \times 10 = 5.0 \text{ N}$. The tension is $T = F_b - W = 5.0 \text{ N} - 4.0 \text{ N} = 1.0 \text{ N}$.
16. A 2.0 kg block slides down a frictionless incline at 30° to the horizontal. At the bottom, it collides elastically with a 3.0 kg block initially at rest on a horizontal surface with coefficient of kinetic friction $\mu_k = 0.25$. If the incline has a height of 1.8 m, what is the speed of the 3.0 kg block immediately after the collision?
    Answer: 2.4 m/s First, find the speed of the 2.0 kg block at the bottom using energy conservation: $v_1 = \sqrt{2gh} = \sqrt{2(9.8)(1.8)} = 5.94$ m/s. For an elastic collision between masses $m_1$ and $m_2$ where $m_2$ is initially at rest: $v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2(2.0)}{2.0 + 3.0}(5.94) = 2.38 \approx 2.4$ m/s. Choice B incorrectly uses the initial speed without the collision formula. Choice C uses an incorrect collision formula. Choice D incorrectly assumes complete momentum transfer.
17. A 1200 kg car traveling at 25 m/s applies brakes and skids to a stop. The coefficient of kinetic friction between tires and road is 0.8. During braking, a 500 N air resistance force also acts opposite to motion. What distance does the car travel before stopping?
    Answer: 28.4 m because both friction and air resistance oppose motion Total stopping force = kinetic friction + air resistance = $\mu_k mg + F_{air} = 0.8(1200)(9.8) + 500 = 9408 + 500 = 9908$ N. Using work-energy theorem: $\frac{1}{2}mv^2 = F_{total} \times d$, so $d = \frac{mv^2}{2F_{total}} = \frac{1200(25)^2}{2(9908)} = \frac{750000}{19816} = 28.4$ m. Choice A uses only friction force. Choice B ignores air resistance and uses an incorrect friction calculation. Choice D incorrectly assumes air resistance reduces friction rather than adding to the stopping force.
18. A 0.2 kg hockey puck moving at 15 m/s collides with a stationary 0.15 kg puck. After collision, the first puck moves at 8.0 m/s at 30° from its original direction, and the second puck moves at angle θ from the original direction of the first puck. What is the magnitude of the second puck's velocity after collision?
    Answer: 12.1 m/s because momentum conservation requires specific velocity magnitude Use conservation of momentum in x and y directions. Initial: $p_x = 0.2 \times 15 = 3.0$ kg⋅m/s, $p_y = 0$. Final x-component: $p_x = 0.2 \times 8.0 \times \cos(30°) + 0.15 \times v_2 \times \cos(θ) = 1.39 + 0.15v_2\cos(θ)$. Final y-component: $p_y = 0.2 \times 8.0 \times \sin(30°) + 0.15 \times v_2 \times \sin(θ) = 0.8 + 0.15v_2\sin(θ)$. From x: $3.0 = 1.39 + 0.15v_2\cos(θ)$, so $0.15v_2\cos(θ) = 1.61$. From y: $0 = 0.8 + 0.15v_2\sin(θ)$, so $0.15v_2\sin(θ) = -0.8$. Therefore: $v_2 = \sqrt{\frac{(1.61)^2 + (-0.8)^2}{(0.15)^2}} = 12.1$ m/s. Other choices use incorrect momentum conservation applications.
19. A small sphere falls through a viscous fluid and reaches a terminal velocity $v_t$. If the experiment is repeated with a sphere of the same size and material but in a fluid with four times the viscosity, what will be the new terminal velocity?
    Answer: $v_t/4$ At terminal velocity, the net force on the sphere is zero. The downward gravitational force (weight) is balanced by the upward buoyant force and the viscous drag force. Thus, $F_g = F_b + F_d$. The drag force is given by Stokes' Law, $F_d = 6\pi\eta r v_t$. Since $F_g$ and $F_b$ are constant for the same sphere and fluid density, the drag force $F_d$ must also be constant at terminal velocity. This means the product $\eta v_t$ is constant. If the viscosity $\eta$ is multiplied by 4, the terminal velocity $v_t$ must be divided by 4 to keep the product constant.
20. A car travels at a constant speed over a circular hump-backed bridge of radius $r$. What is the maximum speed the car can have at the highest point of the bridge without losing contact with the road?
    Answer: $\sqrt{gr}$ At the highest point of the bridge, the net force on the car provides the centripetal force required for circular motion. The forces acting on the car are gravity ($mg$, downwards) and the normal force from the road ($F_N$, upwards). The net force is $mg - F_N$, directed towards the center of the circle (downwards). So, $mg - F_N = mv^2/r$. The car loses contact with the road when the normal force $F_N$ becomes zero. Setting $F_N = 0$, we get $mg = mv_{max}^2/r$. Solving for $v_{max}$ gives $v_{max} = \sqrt{gr}$.

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