Impulse

IMPULSE The momentum alteration experienced by a system resulting from the application of a force (F) over an infinitesimal period is termed impulse (1), which is imparted to the system by the force. vec I = int vec F dt= Delta vec p = vec p t - vec p i The impulse exerted on an object within a specific time interval corresponds to the area beneath the force-versus-time (F - t) graph within that interval. = vec f = int vec F dt= Delta vec p = vec p t - vec p 1 Note: If a force is capable of significantly altering momentum even within an infinitesimally small duration, it qualifies as an impulsive force. Ex: Normal reaction, friction force, tension force, etc. Ex. A ball of mass m travelling with velocity 2i + 3j receives an impulse 3mi. What is the velocity of the ball immediately afterwards? Sol Given, Mass of the ball Initial velocity, u = 2i + 3j Impulse, vec j = - 3mi Let vec v the final velocity of the ball Here, Initial momentum, vec p_{i} = m(2i + 3j) Final momentum, p_{t} = mv We know that the imparted on a system by a force is equal to the change in momentum of the system. -3mi=m vec v - m(2i + 3j) vec I = Delta vec p = vec p f - vec p i - 3i + (2i + 3j) = vec v overline v = - i + 3j = m \ Ex. A bullet of mass 10 ^ - 3 rfloor strikes an obstacle and moves 60 deg to its original direction. If its speed also changes from 20m * s ^ - 1 to 10 m * s ^ - 1 find the magnitude of impulse acting on the bullet. Sol. Given, Mass of the bullet, m = 10 ^ - 3 * kg Initial speed, v_{i} = 20m * s ^ - 1 Final speed, v_{l} = 10m * s ^ - 1 20m * s ^ - 1 y X 120 deg 60 deg 10m * s ^ - 1 i_{f} = 10cos 60 deg * (i) + 10sin 60 deg * (- j) = (5i - 5sqrt(3) * j) * m * s ^ - 1 vec lv = vec v t - vec v_{t} = \{(5i - 5sqrt(3) * j) - 20i\} Delta vec v =(-15i-5 sqrt 3 overline j ) m * s ^ - 1 vec v_{1} = 20im * s ^ - 1 X 20m * s ^ - 1 120 60 10m * s ^ - 1 10sin 60 deg We know that the impulse imparted by a force on a system is equal to the change in momentum. of that system. vec J =m vec v t -m vec v 1 =m Delta vec v vec I = Delta vec p = vec p t - vec p i = m(- 15i - 5sqrt(3) * j) = 10 ^ - 3 * (- 15i - 5sqrt(3) * j) vec |J |= 10 ^ - 3 * sqrt((- 15) ^ 2 + (- 5sqrt(3)) ^ 2) = 10 ^ - 3 * sqrt(300) = 10 ^ - 2 * sqrt(3) * Ns | vec J |= sqrt(3) * 10 ^ - 2 * Ns Ex. Two identical blocks A and B of mass m that are connected by a massless string are placed on a frictionless horizontal plane. A bullet having the same mass and moving with speed u strikes block B as shown in the figure. If the bullet gets embedded into the block B, then find the following: Smooth A B m m The final velocity of blocks A, B, and the bullet (a) (b) The impulse on block A due to the tension in the string (c) The impulse on the bullet due to the normal force (d) The impulse on block B due to the normal force Sol Given, Mass of the bullet = m Mass of each block Speed of the bullet = u (a) Final velocity of blocks A, B, and the bullet Consider the two blocks and the bullet together as a system. Smooth A Let vec v | the final velocity of the system. Since the net external force acting on the system along the x-direction is zero, the momentum of the system is conserved along the x-direction. 0 + 0 + mu = (m + m + m) * V Sigma( vec F ext ) * = vec 0 Rightarrow( vec p l ) * =( vec p l ) * V = u/3 (b) Impulse on block A due to the tension in the string Let I_{tau} the impulse acting on block A due to the tension force. Then vec I_{T} = m vec u 3 vec I_{T} =m( vec V f - vec V i )=m( vec u 3 - vec 0 ) | vec I T |= mu/3 (c) Impulse on the bullet due to the normal force Let vec j_{u} be the impulse acting on the bullet due to the normal force from block B. Then, | hat J B |= (2mu)/3 along negative x-axis) vec j_{B} =m( vec V f - vec V i )=m( vec u 3 - vec u ) vec I_{B} =- 2m vec u 3 (d) Impulse on block B normal force By Newton's third law of motion, the impulse on block B will be negative of the impulse on the bullet. Ex. A particle of mass 2 kg is initially at rest. A force starts acting on the particle in one direction whose magnitude changes with time. The force vs time graph is shown in the figure. Find the speed of the particle at the end of 10 s. 20 F(N) 10 246 10(s) PIN 20 Sol. Given, Mass of the particle, m = 21 Initial speed of particle, v_{i} = 0m * s ^ - 1 Let vi be the final speed of the particle after 10 s. We know that the area under the force vs time graph gives us impulse. Thus, 10 However, impulse imparted is equal to change in the momentum. 100 = 2v_{t} I = m*Delta*v = m(v_{f} + 0) v_{f} = 50m * s ^ - 1 I = 100Ns I = A_{1} + A_{2} + A_{3} + A_{4} I = (1/2 * 2 * 10) + (2 * 10) + (1/2 * (10 + 20) * 2) + (1/2 * 4 * 20) Ex. In the figure shown, the heavy ball of mass 2m rests on the horizontal surface and the lighter ball of mass m is dropped from a height h > 2l At the instant, the string gets taut. What is the upward velocity of the heavy ball? 2m 10 h > 21 A_{i} Sol When the lighter ball of mass m is dropped from the initial position, it falls freely for length 21 with the final velocity being v. At this instant, the string gets taut and the velocity of the system becomes v_{1} as shown in the figure. 2m h > 21 2m Iv m 2 moly, v_{1} 1o2m Choosing point 2 as the datum and applying principle of conservation of mechanical energy between points 1 and 2, we get the following: K_{1} + U_{1} = K_{2} + U_{2} 0 + mg(2l) = 1/2 * m * v ^ 2 + 0 ... (1) As soon as the lighter ball reaches the point 2, both the balls experience an impulse Jr due to the v = 2sqrt(gl) tension force. For the heavier ball, we get the following: I_{T} = 2m(v_{1} + 0) For the lighter ball, we get the following: I_{T} = m(- v_{1} - (- v)) By comparing equations (2) and (3), we get the following: 2m*v_{1} = m(v - v_{1}) ... (2) ...(3) I_{r} = m(v - v_{1}) I_{T} = 2m*v_{1} v_{1} = y/3 Putting the value of from equation (1), we get the following: v_{1} = (2sqrt(yl))/3 Impulsive force: An impulsive force, characterized by its higher magnitude and brief duration, is known for its ability to swiftly alter the momentum of an object within a short span of time. It's important to note that the distinction between impulsive and non-impulsive forces is relative and lacks a precise boundary. Note: Typically, forces involved in collisions exhibit impulsive characteristics. Due to their brief application period, there is minimal displacement of particles, resulting in limited motion. 1. Gravitational force and spring force consistently lack impulsive properties. 2. The nature of normal, tension, and friction forces varies depending on the situation. 3. An impulsive force can only be counteracted by another impulsive force. 1. Impulsive Normal: During a collision, the normal forces acting on the collision surface are invariably impulsive. Ex: M_{i} ^ - 1 m_{1} I M 1 N_{1} = Impulsive; N_{epsilon} = Non-impulsive Both normal are Impulsive N_{1}, N_{3} = Impulsive; N_{2} = non-impulsive Both normal are Impulsive dot \mathfrak{N} 2. Impulsive Friction: If the normal force exerted between the two objects is impulsive, then the friction force between them will also exhibit impulsive characteristics. ↓ N_{2} N_{1} N_{2} N_{g} N_{2} Friction at both surfaces is impulsive Friction due to N_{2} | is non-impulsive and due to N_{3} And N_{1} are impulsive 3. Impulsive Tensions: When a string experiences a sudden jerk, equal and opposite tensions arise simultaneously at each end. As a result, equal and opposite impulses are exerted on the bodies connected to the string in the direction of the string. This scenario presents two distinct cases that need to be examined. (a) One end of the string is fixed: The impulse exerted at the fixed end of the string does not alter the momentum of the fixed object located there. Conversely, the object connected to the free end experiences a momentum change aligned with the direction of the string. Additionally, momentum remains unaltered in a direction perpendicular to the string, where impulsive forces are absent. (b) Both ends of the string attached to movable objects: T Tis Impulsive In this scenario, equal and opposite impulses affect the two objects, resulting in equivalent alterations in momentum. Consequently, the total momentum of the system remains constant, even though each object experiences a change in momentum aligned with the string's direction. However, perpendicular to the string, no impulse is applied, thus leaving the momentum of each particle in this direction. unaffected. T is non-impulsive B For a rod, tension consistently exhibits impulsive characteristics, whereas for a spring, tension consistently lacks impulsive properties. Tis non-impulsive All normal are impulsive but tension Tis impulsive only for the ball A Ex. Two identical blocks, labeled A and B, are connected by a massless string and situated on a frictionless horizontal surface. A bullet with the same mass, traveling at speed u, collides with block B from behind, as illustrated. If the bullet becomes embedded in block B, determine: C A B ................................. (a) The velocity of A,B,C after collision. m m (b) Impulse on A due to tension in the string (c) Impulse on C due to normal force of collision. (d) Impulse on B due to normal force of collision. Sol (a) By Conservation of linear momentum v = = (b) Tdt= mu (c) Ndt = m(-u)= (d) (N-T)dt=Ndt - Tdt = JNdt=