Physics Chapter 9: Linear Momentum

The kinetic energy of the particle is given by K = 1/2mv², where v is the speed of the particle. We know that the momentum of the particle is given by p = mv. We can rearrange this to find that v = p/m. Substituting this expression for v in the kinetic energy equation, we get K = 1/2m(p/m)², which simplifies to K = p²/2m. This is the desired result.

Solving the equation in part (a) for the momentum, we get p = √[2mK]. This expresses the magnitude of the particle's momentum in terms of its kinetic energy and mass.

The momentum is given by p = mv. Using the given velocity, the x-component of momentum is px = (3.00 kg)(3.00 m/s) = 9.00 kg·m/s. The y-component of momentum is py = (3.00 kg)(-4.00 m/s) = -12.0 kg·m/s. The negative sign indicates that the y-component of momentum is in the negative y direction.

<p>The magnitude of the momentum is given by <em>p = √(px² + py²) = √[(9.00 kg·m/s)² + (-12.0 kg·m/s)²] = 15.0 kg·m/s</em>. The direction of the momentum is given the angle <em>θ = tan⁻¹ (py/px) = tan⁻¹ (-12.0 kg·m/s / 9.00 kg·m/s) = -53.1°</em>. The momentum is directed 53.1° to the right of the negative y-axis (or 36.9° below the positive x-axis).</p> Signup and view all the answers

<p>The average vector force exerted by the ball on the bat is <em>F = Δp/Δt</em>. We can calculate the change in momentum as <em>Δp = mvf - mvi</em>. The initial velocity in the x-direction is 45.0 m/s, and the final velocity in the y-direction is 55.0 m/s. Since the ball is in contact with the bat for 2.00 ms which is equal to 0.002 s, the average force exerted by the ball on the bat is <em>F = (0.145 kg)(55.0 j m/s - 45.0 i m/s)/0.002 s = 7 250 i N - 4 037.5 j N</em>.</p> Signup and view all the answers

<p>Since the total momentum is zero, there will be an equal but opposite momentum for the girl. So, the girl will move eastward at the velocity of 2.90 m/s, as <em>m1v1 = m2v2</em>.</p> Signup and view all the answers

<p>The potential energy converted to mechanical energy is given by <em>1/2(m1v1² + m2v2²) = 1/2(65.0 kg)(2.90 m/s)² + 1/2(40.0 kg)(2.90 m/s)² = 320 J</em>.</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>The initial momentum of the system was zero. This is because both blocks were at rest, and the initial momentum of the system is given by <em>m1vi1 + m2vi2</em>. Since the initial momentum of the system is zero, and the momentum is conserved, the final momentum of the system is also zero. This means <em>m1vf1 + m2vf2 = 0</em>. The final momentum of the block with mass <em>3m</em> is <em>3m</em> * 2.00 m/s*. Therefore, the momentum of the block with mass <em>m</em> is <em>m</em> * (-6.00 m/s)* and the final velocity of the block with mass <em>m</em> is <em>- 6.00 m/s</em>. The negative sign indicates that the block of mass <em>m</em> is moving to the left.</p> Signup and view all the answers

<p>The original elastic potential energy stored in the spring is equal to the sum of the kinetic energies of the two blocks after the blocks are released. This is because after the cord is burned, the energy stored in the spring is released, and the kinetic energies of the two blocks are obtained. So, the original elastic potential energy is given by <em>KEtotal = 1/2m1v1² + 1/2m2v2² = 1/2(0.350 kg)(-6.00 m/s)² + 1/2(3</em>0.350 kg)(2.00 m/s)² = 7.35 J*.</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>The recoil speed is the speed at which the Earth would move in response to you jumping. It is given by <em>vEarth = (mYou/mEarth)vYou</em>. Since the mass of the Earth is very large compared to your mass, the recoil speed is very small and can be considered negligible.</p> Signup and view all the answers

<p>Python = General-purpose programming JavaScript = Client-side scripting for web applications SQL = Database queries CSS = Styling web pages</p> Signup and view all the answers

<p>The potential energy stored in the spring is given by <em>U = 1/2kx²</em>. The potential energy is converted into kinetic energy of the glider as the glider is released. Hence, <em>1/2mv² = 1/2kx²</em>. Solving this equation for the speed of the glider, we get <em>v = x(k/m)1/2</em>.</p> Signup and view all the answers

<p>The impulse imparted to the glider is given by <em>I = Δp = mv - 0</em> or, <em>I = mv</em>. Substituting the velocity of the glider from part (a) into the impulse equation, we get <em>I = mx(k/m)¹/² = x(km)¹/²</em>.</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>The force required to hold the child is given by <em>F = Δp/Δt = mΔv/Δt</em>. Using the given values, we get <em>F = (12kg)(60mi/h * 0.447 m/s/1mi/h)/0.1s = 3 200 N</em>. The force needed to hold the child is very high, and it could be impossible for a person to exert such a strong force. It clearly shows that it is not safe to hold a child on their lap, and using a car seat for the child is highly recommended.</p> Signup and view all the answers

<p>The time it takes for the car to stop is given by <em>t = Δv/a</em>. Using the initial velocity and distance covered, we get <em>t = 25.0 m/s / (25.0 m/s)²/ (2 * 1.20 m) = 0.192 s</em>. The collision lasts for 0.192 seconds.</p> Signup and view all the answers

<p>The average force exerted on the car is given by <em>F = mΔv/Δt = (1 400 kg)(25.0 m/s)/0.192 s = 1.82 × 10⁵ N</em>. The average force exerted on the car is 1.82 × 10⁵ N.</p> Signup and view all the answers

<p>The acceleration of the car is given by <em>a = Δv/Δt = (25.0 m/s)/0.192s = 130 m/s²</em>. The magnitude of the acceleration is about 13 times the acceleration due to gravity, which is 9.8 m/s².</p> Signup and view all the answers

<p>The impulse of the force is given by <em>I = Δp = FΔt</em>. The impulse is the change in momentum. The impulse is given by the area under the force versus time graph. Finding the area of the triangular shape from <em>t = 0</em> to <em>t = 3</em> is <em>1/2(3 sec)(3 N) = 4.5 N·s</em>. The area of the rectangular is <em>1(3 N) = 3 N·s</em>. So, the total impulse is given by <em>(4.5 N·s + 3 N·s) = 7.5 N·s</em>.</p> Signup and view all the answers

<p>The final velocity of the particle is given by <em>vf = vi + Δv</em>. The final velocity is given by the change in the momentum and the initial velocity. We can calculate the change in velocity from the impulse as <em>Δv = I/m = 7.5 N·s / 2.50 kg = 3.00 m/s</em>. Therefore, the final velocity of the particle is <em>vf = 0 + 3.00 m/s = 3.00 m/s</em>.</p> Signup and view all the answers

<p>The final velocity of the particle is given by <em>vf = vi + Δv</em>. The impulse is independent of the initial velocity, so the change in velocity is still <em>Δv = 3.00 m/s</em>. The final velocity is, therefore, <em>vf = -2.00 m/s + 3.00 m/s = 1.00 m/s</em>.</p> Signup and view all the answers

<p>The average force exerted on the particle is given by <em>F = I/Δt</em>. The average force is the impulse of the force over the time interval. The average force is <em>F = 7.5 N·s / 5.00 s = 1.5 N</em>.</p> Signup and view all the answers

<p>The scale reads the weight of the bucket plus the weight of the water in the bucket. The mass of the water in the bucket is given by <em>mwater = (0.250 L/s)(1 kg/L)(3 sec) = 0.75 kg</em>. The weight of the water is given by <em>Fwater = (0.75 kg)(9.8 m/s²) = 7.35 N</em>. The scale reads <em>7.35 N + (0.750 kg)(9.8 m/s²) = 14.7 N</em>.</p> Signup and view all the answers

<p>The initial momentum of the system is <em>m1vi1 + m2vi2 = (1 200 kg)(25.0 m/s) + (9 000 kg)(20.0 m/s) = 2.16 × 10⁵ kg·m/s</em>. Since the momentum is conserved, this must be equal to the final momentum <em>m1vf1 + m2vf2</em>. Solving for <em>vf2</em>, we get <em>vf2 = (2.16 × 10⁵ kg·m/s - (1 200 kg)(18.0 m/s)) / 9 000 kg = 20.8 m/s</em>. Therefore, the final velocity of the truck is 20.8 m/s.</p> Signup and view all the answers

<p>The initial kinetic energy of the system is <em>1/2m1v1² + 1/2m2v2² = 1/2(1 200 kg)( 25.0 m/s)² + 1/2(9 000 kg)(20 m/s)² = 2.43 × 10⁶ J</em>. The final kinetic energy of the system is <em>1/2m1vf1² + 1/2m2vf2² = 1/2(1 200 kg)(18 m/s)² + 1/2(9 000 kg)(20.8 m/s)² = 2.23 × 10⁶ J</em>. The change in the mechanical energy is <em>ΔKE = (2.23 × 10⁶ J - 2.43 × 10⁶ J) = -2.0 × 10⁵ J</em>. The mechanical energy of the system is not conserved during the collision.</p> Signup and view all the answers

<p>The loss in the mechanical energy is converted into other forms of energy, such as sound, heat, and deformation of the car and the truck. The collision is not perfectly elastic, so the mechanical energy is not conserved. The initial kinetic energy is converted into other forms of energy and the final kinetic energy is less than the initial kinetic energy. The initial and final momentum of the system is conserved because there are no external forces acting on the car and truck system.</p> Signup and view all the answers

<p>The initial momentum of the system is <em>m1vi1 + m2vi2 + m3vi3 + m3vi4 = (2.50 × 10⁴ kg)(4.00 m/s) + (2.50 × 10⁴ kg)(2.00 m/s) + (2.50 × 10⁴ kg)(2.00 m/s) + (2.50 × 10⁴ kg)(2.00 m/s) = 2.50 × 10⁵ kg·m/s</em>. The final momentum of the system is <em>m1vf1 + m2vf2 + m3vf3 + m4vf4 = (2.50 × 10⁴ kg + 2.50 × 10⁴ kg + 2.50 × 10⁴ kg + 2.50 × 10⁴ kg)vf = 10⁵ kgvf</em>. Since the momentum is conserved <em>2.50 × 10⁵ kg·m/s = 10⁵ kgvf</em>. Therefore, the final velocity of the four cars is <em>vf = 2.50 m/s</em>.</p> Signup and view all the answers

<p>The initial kinetic energy of the system is <em>KEi = 1/2m1v1² + 1/2(m2v2² + m3v3² + m4v4²) = 1/2(2.50 × 10⁴ kg)(4.00 m/s)² + 1/2((2.50 × 10⁴ kg)(2.00 m/s)² + (2.50 × 10⁴ kg)(2.00 m/s)² + (2.50 × 10⁴ kg)(2.00 m/s)²) = 3.75 × 10⁵ J</em>. The final kinetic energy of the system is <em>KEf = 1/2(2.50 × 10⁴ kg + 2.50 × 10⁴ kg + 2.50 × 10⁴ kg + 2.50 × 10⁴ kg)(2.50 m/s)² = 1.56 × 10⁵ J</em>. The decrease in mechanical energy is given by *ΔKE = (KEi - KEf) = 3.75 × 10⁵ J - 1.56 × 10⁵ J = 2.</p> Signup and view all the answers

<p>Since the system is isolated and momentum is conserved, the girl will move in the opposite direction with a velocity that ensures the total momentum of the boy-girl system remains zero. The velocity of the girl can be calculated using the conservation of momentum equation: <em>m_b</em> <em>v_b</em> + * m_g* <em>v_g</em> = 0, where <em>m_b</em> is the boy's mass, <em>v_b</em> is the boy's velocity, <em>m_g</em> is the girl's mass, and <em>v_g</em> is the girl's velocity. Solving for <em>v_g</em> gives <em>v_g = - (m_b</em> <em>v_b</em>)/ <em>m_g</em> = - (65.0 kg * 2.90 m/s) / 40.0 kg = - 4.49 m/s*. The girl will move eastward with a velocity of 4.49 m/s.</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>The potential energy stored in the spring is <em>U = (1/2)kx²</em>. When the spring is released, this potential energy is converted into kinetic energy of the glider, <em>(1/2)mv²</em>. Setting these two expressions equal gives <em>(1/2)kx² = (1/2)mv²</em>. Solving for <em>v</em> yields: <em>v = x√(k/m)</em>.</p> Signup and view all the answers

<p>Impulse is defined as the change in momentum, <em>I = Δp = m(v_f - v_i)</em>. Since the glider starts from rest, <em>v_i = 0</em>. Using the expression for final velocity we derived previously, <em>v_f = x√(k/m)</em>, we can substitute it into the impulse equation giving <em>I = m(x√(k/m)) = x√(km)</em>.</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>(a) The collision lasts for a time interval of Δt = Δx/v = 1.20 m / 25.0 m/s = 0.048 s. (b) The magnitude of the average force acting on the car is F = m(v_f - v_i)/Δt = (1 400 kg)(0 - 25.0 m/s) / 0.048 s = 7.29 x 10⁵ N. (c) The magnitude of the acceleration of the car is a = (v_f - v_i)/Δt = (0 - 25.0 m/s) / 0.048 s = 5.21 x 10² m/s² or 53.1 g, where g is the acceleration due to gravity.</p> Signup and view all the answers

<p>(a) The impulse is equal to the area under the force-time curve. Calculating the area under the curve gives <em>I = (1/2)(4 N)(2 s) + (3 N)(1 s) + (1/2)(1 N)(2 s) = 12 N·s</em>. (b) The final velocity of the particle when initially at rest can be calculated using the impulse-momentum theorem: <em>I = m(v_f - v_i)</em>. Substituting the values gives <em>12 N·s = 2.50 kg (v_f - 0)</em>, giving <em>v_f = 4.80 m/s</em>. (c) If the initial velocity is -2.00 m/s, then the final velocity is <em>v_f = I/m + v_i = 12 N·s / 2.50 kg - 2.00 m/s = 2.80 m/s</em>. (d) The average force acting on the particle is the total impulse divided by the time interval. Therefore, <em>F_avg = I/Δt = 12 N·s / 5.00 s = 2.40 N</em>.</p> Signup and view all the answers

<p>(a) Using the conservation of momentum, we can find the velocity of the truck after the collision: <em>m_car * v_car_i + m_truck * v_truck_i = m_car * v_car_f + m_truck * v_truck_f</em>. Solving for <em>v_truck_f</em> gives: <em>v_truck_f = (m_car * v_car_i + m_truck * v_truck_i – m_car * v_car_f) / m_truck = (1 200 kg * 25.0 m/s + 9 000 kg * 20.0 m/s - 1 200 kg * 18.0 m/s) / 9 000 kg = 20.4 m/s</em>. (b) The change in mechanical energy is <em>ΔKE = KE_f - KE_i = (1/2)m_car</em> <em>v_car_f² + (1/2)m_truck</em> <em>v_truck_f² - (1/2)m_car</em> <em>v_car_i² - (1/2)m_truck</em> <em>v_truck_i² = (1/2)(1 200 kg)(18.0 m/s)² + (1/2)(9 000 kg)(20.4 m/s)² - (1/2)(1 200 kg)(25.0 m/s)² - (1/2)(9 000 kg)(20.0 m/s)² = -7.20 x 10⁴ J</em>. (c) This negative change in mechanical energy is due to the inelastic nature of the collision. Some of the initial kinetic energy was transformed into other forms of energy, such as sound energy and heat energy, which are dissipated in the environment.</p> Signup and view all the answers

<p>(a) Using the conservation of momentum, the final speed of the four cars can be calculated by: <em>m1</em> *v1_i + <em>m2</em> <em>v2_i = (m1 + m2) * v_f</em>. where <em>m1</em> = 2.50 x 10⁴ kg, *v1_i = 4.00 m/s, <em>m2</em> = 3(2.50 x 10⁴ kg) = 7.50 x 10⁴ kg, and <em>v2_i = 2.00 m/s</em>. Solving for <em>v_f</em> gives <em>v_f = (m1 * v1_i + m2 * v2_i) / (m1 + m2) = (2.50 x 10⁴ kg * 4.00 m/s + 7.50 x 10⁴ kg * 2.00 m/s) / (2.50 x 10⁴ kg + 7.50 x 10⁴ kg) = 2.67 m/s</em>. (b) The decrease in mechanical energy is <em>ΔKE = KE_i - KE_f = (1/2)m1</em> <em>v1_i² + (1/2)m2</em> <em>v2_i² - (1/2)(m1 + m2)</em> <em>v_f² = (1/2)(2.50 x 10⁴ kg)(4.00 m/s)² + (1/2)(7.50 x 10⁴ kg)(2.00 m/s)² - (1/2)(2.50 x 10⁴ kg + 7.50 x 10⁴ kg)(2.67 m/s)² = -1.67 x 10⁵ J</em>. This negative change in mechanical energy is due to the inelastic nature of the collision, where some of the kinetic energy was converted to other forms of energy, such as heat and sound.</p> Signup and view all the answers

<p>(a) Since the momentum of the system is conserved, the total initial momentum of the four cars must equal the total final momentum. We can write this equation as: <em>m_1</em> *v_i + <em>m_2</em> *v_i + <em>m_3</em> *v_i + <em>m_4</em> *v_i = <em>m_1</em> *v_1_f + <em>m_2</em> *v_2_f + <em>m_3</em> <em>v_3_f</em>. Since all the cars have the same mass (2.50 x 10⁴ kg), we can simplify this to <em>4mv_i = mv_1_f + 3mv_2_f</em>. Substituting the values for the final velocities we obtain <em>4v_i = 4.00 m/s + 3(2.00 m/s)</em>. Solving for <em>v_i</em> gives: <em>v_i = 2.50 m/s</em>. (b) The potential energy of the actor did not change because the actor exerted a force on the front car, pushing it, but the actor himself was already moving at the initial speed of 2.50 m/s. This means that the total kinetic energy of the actor did not change during the pushing process. (c) The process described here is essentially the reverse of the process in Problem 13. In Problem 13, the coupling of the four cars resulted in a decrease in kinetic energy, while in this problem, the uncoupling and pushing of the front car results in an increase in kinetic energy. Both problems illustrate the principle of conservation of momentum and energy, but they involve different scenarios and emphasize different aspects of these principles.</p> Signup and view all the answers

<p>(a) The total momentum of the system (car + truck) is conserved in the collision. We can write this as <em>m1</em> <em>v1_i + <em>m2</em> <em>v2_i = (<em>m1 + <em>m2</em>) * v_f</em>, where <em>m1</em> = <em>m</em>, <em>v1_i</em> = <em>v1</em>, <em>m2</em> = 2</em>m</em>, <em>v2_i</em> = <em>v2</em>, and <em>v_f</em> is the final velocity of the combined system. Solving for <em>v_f</em> gives: <em>v_f = (<em>m1</em> <em>v1_i + <em>m2</em> <em>v2_i) / (<em>m1 + <em>m2</em>) = (<em>m</em> <em>v1</em> + 2</em>m</em> <em>v2</em>) / (3</em>m</em>) = (v1 + 2v2) / 3*. (b) The change in kinetic energy is *ΔKE = KE_f - KE_i = (1/2)(*m1 + <em>m2</em>) <em>v_f² - (1/2)<em>m1</em> <em>v1_i² - (1/2)<em>m2</em> <em>v2_i² = (1/2)(3</em>m</em>)(v1 + 2v2)² / 9 - (1/2)<em>m</em> <em>v1² - (1/2)(2</em>m</em>) <em>v2² = -m(v1² + 2v1v2 + 4v2²) / 18</em>. The negative sign indicates that the kinetic energy of the system decreased during the collision.</p> Signup and view all the answers

<p>The depth to which the bullet penetrates the block depends on the work done by the bullet's force. The work done is equal to the change in kinetic energy of the bullet. Since the bullet is fired at the same initial speed in both cases, and the work done is equal to the change in kinetic energy, which is the same in both cases, the penetration depth will be the same in both cases, 8.00 cm.</p> Signup and view all the answers

<p>(a) The final velocity of the basketball when it hits the ground can be calculated using the equation <em>v² = 2gh</em>: <em>v = √(2gh) = √(2 x 9.8 m/s² x 1.20 m) = 4.85 m/s</em>. (b) Since the collision is perfectly elastic, the kinetic energy of the system (basketball + tennis ball) is conserved. Let <em>v1</em> and <em>v2</em> be the velocities of the basketball and the tennis ball, respectively, before the collision, and <em>v1’</em> and <em>v2’</em> be their velocities after the collision. We can apply the conservation of energy equation: <em>(1/2)m1v1² + (1/2)m2v2² = (1/2)m1v1’² + (1/2)m2v2’²</em>. Since both the tennis ball and the basketball have the same mass, <em>m1 = m2 = m</em>, and since the collision is elastic, the speeds of the two balls are swapped after the collision. <em>v2 = 4.85 m/s</em> and <em>v1 = -4.85 m/s</em>. Therefore, <em>v2’ = 4.85 m/s</em> and *v1’ = -4.85 m/s.</p> Signup and view all the answers