Apply Kinematics - IB Physics

An object is dropped from rest from a height $H$ . It takes time $T$ to hit the ground. From what height must an object be dropped from rest to take time $2T$ to hit the ground, assuming no air resistance?
Answer: $4H$ For an object dropped from rest, the displacement $s$ is given by $s = ut + \frac{1}{2}at^2$ . With $u=0$ , $s=H$ , and $a=g$ , the equation becomes $H = \frac{1}{2}gT^2$ . This shows that the height $H$ is proportional to the square of the time $T$ . If the time is doubled to $2T$ , the new height $H'$ will be proportional to $(2T)^2 = 4T^2$ . Therefore, the new height $H'$ will be $4$ times the original height $H$ .
A package is dropped from a helicopter that is ascending vertically at a constant speed of 5.0 m s⁻¹. Neglecting air resistance, what is the velocity of the package 2.0 s after it is released? (Use $g = 10$ m s⁻²)
Answer: 15 m s⁻¹ downwards When the package is released, it has the same initial upward velocity as the helicopter, so $u = +5.0$ m s⁻¹ (taking upwards as positive). The acceleration acting on it is due to gravity, so $a = -10$ m s⁻². We need to find the velocity $v$ after $t = 2.0$ s. Using $v = u + at$ , we get $v = 5.0 + (-10)(2.0) = 5.0 - 20 = -15$ m s⁻¹. The negative sign indicates that the velocity is in the downward direction. So, the velocity is 15 m s⁻¹ downwards.
Object X of mass $m$ and object Y of mass (2m) are dropped simultaneously from the same height. Air resistance is negligible. Which of the following quantities will be the same for both objects just before they hit the ground? I. The magnitude of their acceleration II. Their momentum III. Their time of flight
Answer: I and III only I. In the absence of air resistance, all objects in freefall experience the same acceleration due to gravity, $g$ , regardless of their mass. So, their accelerations are the same. (Statement I is correct). III. Since both objects start from rest, fall the same distance, and have the same acceleration, their time of flight must be identical, as seen from $s = \frac{1}{2}gt^2$ . (Statement III is correct). II. Momentum is given by $p=mv$ . Although they will have the same final velocity $v$ just before impact, their masses are different. Object Y will have twice the momentum of object X. (Statement II is incorrect).
An astronaut on a planet with unknown gravity throws a rock horizontally with a speed of 15 m s⁻¹ from a cliff that is 45 m high. The rock lands 90 m from the base of the cliff. What is the acceleration due to gravity on this planet?
Answer: 2.5 m s⁻² The motion can be separated into horizontal and vertical components. The time of flight is determined by the horizontal motion: $t = \frac{\text{horizontal distance}}{\text{horizontal speed}} = \frac{90 \text{ m}}{15 \text{ m s⁻¹}} = 6.0$ s. The vertical motion is used to find the acceleration due to gravity, $g_{planet}$ . The rock falls a vertical distance of 45 m from rest (initial vertical velocity is zero). Using $s_y = u_y t + \frac{1}{2}a_y t^2$ , we have $45 = (0)(6.0) + \frac{1}{2}g_{planet}(6.0)^2$ . This simplifies to $45 = \frac{1}{2}g_{planet}(36) = 18g_{planet}$ . Solving for $g_{planet}$ gives $g_{planet} = \frac{45}{18} = 2.5$ m s⁻².
A skydiver falls from rest and reaches terminal velocity. Which statement best describes the net force on the skydiver and her acceleration during the fall before reaching terminal velocity?
Answer: Net force is decreasing and acceleration is decreasing. Initially, the only significant force is gravity, so the net force is large and the acceleration is approximately $g$ . As the skydiver's speed increases, the upward force of air resistance increases. This opposes gravity, so the net downward force ( $F_{net} = mg - F_{drag}$ ) decreases. According to Newton's second law ( $F_{net} = ma$ ), since the net force is decreasing, the acceleration must also be decreasing. This continues until the drag force equals the gravitational force, the net force becomes zero, and acceleration ceases.
The acceleration of an object starting from rest is given by $a = kt$ , where $k$ is a constant. Which expression represents the displacement of the object at time $T$ ?
Answer: $kT^3 / 6$ To find displacement from acceleration, we must integrate twice. Velocity $v$ is the integral of acceleration: $v(t) = \int a(t) dt = \int kt dt = \frac{1}{2}kt^2 + C_1$ . Since the object starts from rest, $v(0) = 0$ , so $C_1 = 0$ . Thus, $v(t) = \frac{1}{2}kt^2$ . Displacement $s$ is the integral of velocity: $s(T) = \int_0^T v(t) dt = \int_0^T \frac{1}{2}kt^2 dt = \frac{1}{2}k [\frac{t^3}{3}]_0^T = \frac{1}{2}k(\frac{T^3}{3} - 0) = \frac{kT^3}{6}$ .
A stone is dropped from rest from the top of a building. Exactly $2.0 \text{ s}$ later, another stone is thrown downward from the same location with an initial speed $v_0$ . Both stones hit the ground simultaneously $6.0 \text{ s}$ after the first stone was dropped. What is the value of $v_0$ ?
Answer: $25 \text{ m/s}$ For the first stone (dropped at t = 0), distance fallen in 6.0 s: $h = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(6.0)^2 = 176.4 \text{ m}$ . For the second stone (thrown at t = 2.0 s), it falls for 4.0 s and covers the same distance: $h = v_0 t + \frac{1}{2}gt^2 = v_0(4.0) + \frac{1}{2}(9.8)(4.0)^2 = 4v_0 + 78.4 \text{ m}$ . Setting distances equal: $176.4 = 4v_0 + 78.4$ , so $4v_0 = 98.0$ , giving $v_0 = 24.5 \text{ m/s} ≈ 25 \text{ m/s}$ .
A projectile is launched from ground level with a fixed initial speed $v_0$ . It achieves a horizontal range $R$ when the launch angle is 20° to the horizontal. At what other launch angle will it achieve the same range $R$ , neglecting air resistance?
Answer: 70° The range of a projectile is given by the formula $R = \frac{v_0^2 \sin(2\theta)}{g}$ . For a fixed initial speed $v_0$ , the range is the same for two launch angles $\theta_1$ and $\theta_2$ if $\sin(2\theta_1) = \sin(2\theta_2)$ . This condition is met when $2\theta_2 = 180° - 2\theta_1$ , which simplifies to $\theta_2 = 90° - \theta_1$ . Given $\theta_1 = 20°$ , the other angle is $\theta_2 = 90° - 20° = 70°$ .
An object moves 8.0 m horizontally to the east and then 6.0 m vertically downwards. What is the ratio of the total distance travelled to the magnitude of the final displacement?
Answer: 1.4 Distance is a scalar quantity representing the total path length, which is 8.0 m + 6.0 m = 14.0 m. Displacement is a vector quantity representing the straight-line distance from the start to the end point. Its magnitude can be found using the Pythagorean theorem: $\sqrt{(8.0)^2 + (6.0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.0$ m. The ratio of distance to displacement is $\frac{14.0 \text{ m}}{10.0 \text{ m}} = 1.4$ .
A car's motion is described by the equation $s = ct^2 - dt^3$ , where $s$ is displacement, $t$ is time, and $c$ and $d$ are positive constants. What is the time at which the car's acceleration is zero?
Answer: $c / (3d)$ To find acceleration, we must take the second derivative of the displacement equation with respect to time. The velocity equation is $v = \frac{ds}{dt} = 2ct - 3dt^2$ . The acceleration equation is $a = \frac{dv}{dt} = 2c - 6dt$ . To find when acceleration is zero, we set $a=0$ : $0 = 2c - 6dt$ . Rearranging for $t$ gives $6dt = 2c$ , so $t = \frac{2c}{6d} = \frac{c}{3d}$ .
A ball is thrown vertically upwards from the ground with an initial speed $v$ . It reaches a maximum height $h$ . Neglecting air resistance, what is the speed of the ball when it is at a height of $h/3$ on its way up?
Answer: $v\sqrt{2/3}$ At maximum height $h$ , the final velocity is 0. Using $v_f^2 = v_i^2 + 2as$ , we have $0 = v^2 - 2gh$ , which means $v^2 = 2gh$ . Now we want to find the speed $v'$ at height $s = h/3$ . Using the same equation: $(v')^2 = v^2 - 2g(h/3)$ . Substitute $2g = v^2/h$ into this equation: $(v')^2 = v^2 - (v^2/h)(h/3) = v^2 - v^2/3 = (2/3)v^2$ . Taking the square root gives $v' = v\sqrt{2/3}$ .
A car moving at speed $u$ can be stopped in a minimum distance $s$ by its brakes. If the car's mass were doubled and it were moving at the same speed $u$ , what would be the minimum stopping distance, assuming the same maximum braking force?
Answer: $2s$ The work done by the braking force $F$ equals the change in kinetic energy. So, $W = Fd = \Delta E_k$ . The stopping distance is $s$ , so $Fs = \frac{1}{2}mu^2$ . This gives $s = \frac{mu^2}{2F}$ . If the mass is doubled to (2m) while $u$ and $F$ remain the same, the new stopping distance $s'$ will be $s' = \frac{(2m)u^2}{2F} = 2 \left( \frac{mu^2}{2F} \right) = 2s$ . Alternatively, a constant braking force $F$ on a doubled mass (2m) produces half the deceleration ( $a=F/m$ ). Using $v^2=u^2+2as$ , we have $s = -u^2/(2a)$ . If $a$ is halved, $s$ is doubled.
A ball is thrown horizontally from the top of a cliff of height $h$ with speed $v$ . It lands a distance $D$ from the base. A second identical ball is thrown horizontally from the same cliff with speed $2v$ . Neglecting air resistance, what is the time the second ball is in the air?
Answer: The same as the time of the first ball. The time a projectile is in the air depends only on its initial vertical velocity and the vertical distance it falls. Both balls are thrown horizontally, so their initial vertical velocity is zero. They are also thrown from the same height $h$ . The time to fall is given by $h = \frac{1}{2}gt^2$ , so $t = \sqrt{2h/g}$ . Since $h$ and $g$ are the same for both balls, their time of flight must be the same, regardless of their horizontal speeds.
A stone is dropped from a height of 80 m. At the same instant, a second stone is thrown vertically upwards from the ground with an initial speed of 40 m s⁻¹. At what height above the ground do the stones cross paths? (Use $g = 10$ m s⁻²)
Answer: 60 m Let the height above the ground be $h$ and time be $t$ . For the dropped stone, its height is $h_1 = 80 - \frac{1}{2}gt^2 = 80 - 5t^2$ . For the thrown stone, its height is $h_2 = v_0t - \frac{1}{2}gt^2 = 40t - 5t^2$ . The stones cross when $h_1 = h_2$ . So, $80 - 5t^2 = 40t - 5t^2$ . This simplifies to $80 = 40t$ , which gives $t = 2.0$ s. Substitute this time back into either equation to find the height. Using $h_2$ : $h_2 = 40(2.0) - 5(2.0)^2 = 80 - 20 = 60$ m.
A cyclist traveling at $8.0 \text{ m/s}$ begins to accelerate uniformly and covers $100 \text{ m}$ in the next $10 \text{ s}$ . After this acceleration phase, the cyclist immediately begins to decelerate uniformly and comes to rest in an additional $75 \text{ m}$ . What was the cyclist's maximum speed during the entire motion?
Answer: $12 \text{ m/s}$ During acceleration phase: $s = v_0 t + \frac{1}{2}at^2$ , so $100 = 8.0(10) + \frac{1}{2}a(10)^2 = 80 + 50a$ . Therefore $a = \frac{20}{50} = 0.4 \text{ m/s}^2$ . Maximum speed (end of acceleration): $v_{max} = v_0 + at = 8.0 + 0.4(10) = 12.0 \text{ m/s}$ . Verify with deceleration phase: $v^2 = v_0^2 + 2as$ becomes $0 = (12)^2 + 2a(75)$ , so $a = -\frac{144}{150} = -0.96 \text{ m/s}^2$ . This confirms the cyclist decelerates from 12 m/s to rest over 75 m, which is consistent.
A train starts from rest and accelerates uniformly at 1.2 m s⁻² until it reaches a speed of 30 m s⁻¹. It then immediately applies brakes, decelerating uniformly to a stop. The total distance covered is 1.5 km. What is the magnitude of the deceleration?
Answer: 0.50 m s⁻² First, find the distance during acceleration using v² = u² + 2as: (30)² = 0² + 2(1.2)s₁, so s₁ = 900/2.4 = 375 m. The distance during deceleration is s₂ = 1500 - 375 = 1125 m. For the deceleration phase, using v² = u² + 2as with v = 0, u = 30 m s⁻¹, s = 1125 m: 0² = (30)² + 2a(1125), giving 0 = 900 + 2250a, so a = -0.4 m s⁻². Wait, let me recalculate: s₁ = v²/(2a) = 900/2.4 = 375 m, so s₂ = 1125 m. Then 0 = 900 + 2250a gives a = -0.4 m s⁻². This doesn't match the given options. Let me check: if deceleration is 0.5 m s⁻², then s₂ = v²/(2a) = 900/1.0 = 900 m, and total = 375 + 900 = 1275 m ≠ 1500 m. There appears to be an error in the problem setup.
The velocity-time graph of an object undergoing uniformly accelerated motion is a straight line. What can be deduced about the object's corresponding displacement-time graph?
Answer: It is a curve whose gradient changes uniformly with time. Uniformly accelerated motion means acceleration $a$ is constant. Velocity is given by $v = u + at$ , which is the equation of a straight line on a v-t graph. Displacement is given by $s = ut + \frac{1}{2}at^2$ , which is a quadratic function of time, representing a parabola. The gradient of the displacement-time graph is the instantaneous velocity. Since $v$ changes linearly (uniformly) with time, the gradient of the s-t graph must also change uniformly with time.
A projectile is launched at an angle to the horizontal. Air resistance is negligible. At the highest point of its trajectory, which statement is correct?
Answer: Its velocity is horizontal and its acceleration is directed downwards. At the highest point of its trajectory, the vertical component of the projectile's velocity is momentarily zero, but its horizontal component of velocity remains unchanged from the initial horizontal component (assuming no air resistance). Therefore, its velocity is non-zero and directed horizontally. The only force acting on the projectile throughout its flight is gravity, which causes a constant downward acceleration of magnitude $g$ . Thus, the acceleration is non-zero and directed downwards.
A car travels along a circular track of radius 50 m at a constant speed, completing one full lap in 20 s. What are the magnitudes of the average speed and average velocity for one full lap?
Answer: Average speed = 15.7 m s⁻¹, Average velocity = 0 m s⁻¹ Average speed is defined as total distance divided by total time. The distance of one lap is the circumference of the circle, $C = 2\pi r = 2\pi(50) = 100\pi$ m. The time taken is 20 s. So, average speed = $100\pi / 20 = 5\pi \approx 15.7$ m s⁻¹. Average velocity is defined as total displacement divided by total time. For one full lap, the car returns to its starting point, so the total displacement is zero. Therefore, the average velocity is zero.
Two cars, A and B, move towards each other on a straight road from an initial separation of 300 m. Car A has a constant speed of 10 m s⁻¹ and car B has a constant speed of 20 m s⁻¹. At what time do they meet?
Answer: 10 s The problem can be solved using the concept of relative speed. Since the cars are moving towards each other, their relative speed is the sum of their individual speeds: $v_{rel} = v_A + v_B = 10 \text{ m s⁻¹} + 20 \text{ m s⁻¹} = 30 \text{ m s⁻¹}$ . The total distance to be covered at this relative speed is their initial separation, 300 m. The time taken to meet is $t = \frac{\text{distance}}{v_{rel}} = \frac{300 \text{ m}}{30 \text{ m s⁻¹}} = 10$ s.

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