EXPERIMENT # 8: IMPULSE AND MOMENTUM PART I: Introduction Newton expressed what we now call his second law of motion, not as �⃗� = ?�⃗�, but in terms of the rate of change of momentum of the object ??...

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EXPERIMENT # 8: IMPULSE AND MOMENTUM PART I: Introduction Newton expressed what we now call his second law of motion, not as �⃗� = ?�⃗�, but in terms of the rate of change of momentum of the object ?? ??⁄ . In this more general and powerful form the law states that when an unbalanced force acts on a body during a finite but short time interval, the change in the object's momentum depends on the product of the force and the time interval for which the force acts. The quantity �⃗�?? is defined as Impulse and the relationship between the change in momentum and the Impulse is sometimes referred to as the Impulse-Momentum Theorem. It states that the integral of the force with respect to time is equal to the change in momentum of the object. When an unbalanced force �⃗�??? acts on an object for a time interval Δt, the momentum of the object will change over this time interval. If ?? and ?? are the initial and final momenta at the start and end of this time interval respectively, then Newton's second law can be written as �⃗�???Δ? = Δ? The product �⃗�??? Δt, when summed over several small time intervals, is the integral of �⃗�??? dt and is defined as the Impulse ?. ? = ∫ �⃗�????? = ?? ?? ?? − ?? Note that impulse is a vector quantity and has the same direction as the change in momentum vector. --------------- Impulse, J = F∆t Unit: Ns F can also be expressed as rate of change of momentum. F = ∆p/∆t  F ∆t = ∆p Impulse = ∆p (Impulse-momentum Theorem) Unit (momentum) = kg m/s PART II: Ballistic Pendulum The ballistic pendulum apparatus consists of a launcher, a projectile, and a pendulum. Attached to the swinging end of the pendulum is a catcher designed to catch the projectile and hold it. The http://en.wikipedia.org/wiki/Newton's_laws_of_motion http://en.wikipedia.org/wiki/Impulse_(physics) projectile can be a metal ball with some mass. The catcher rests directly in front of the launcher, right at the exit point of the projectile (fig. 1). Length of Pendulum is 2 m at angle of 45 degrees. Once the launch mechanism is released, the projectile enters the catcher and the arm swings up, moving past a grooved plastic strip. A ratchet mechanism on the catcher slides past each groove, allowing motion up but not back down. Once the pendulum stops, it stays locked in place recording the maximum height of the swing. Data: Table 1. Masses (g) Ball-I 73.4 Ball-II 110.3 Pendulum 200 Heights to which pendulum rises after collision Ball-I Ball-II Velocities for the ball before collision (cm/s) Ball-I 894.7 Ball-II Elastic Collision: Both momentum and KE conserved. m1v1i + m2v2i = m1v1f +m2v2f (Momentum Conservation) ½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2 v2f2 (Kinetic Energy Conservation) Where i and f represent initial and final values. In-elastic collision: Momentum conserved, KE not conserved. m1v1i + m2v2i = (m1 + m2) Vf (Momentum Conservation)  Find Vf. To find height that the pendulum raises after collision: use the conservation of mechanical energy after the collision. (KE +PE) initial = (KE + PE) final ½ (m1+m2) Vf2 + 0 = 0 + (m1 +m2) g h  Find h. PART III: Moving Block Experiment A pendulum bob with length (L = 3m), vertically positioned is perpendicularly to a Wooden block of mass on a plane surface. The pendulum bob was raised to an angle (θ) to the horizontal and release to strike the wooden block for at least twice. See the table for details. As the bob hit the wooden block, it moved a distance of Xwb at a speed (m/s) as shown in the diagram. (Acceleration due to gravity = 9.8 m/s2) Table 2 : Wooden Block Trial Mass (g) h1 (meters) Xwb (meters) Initial velocity (m/s) Final velocity (m/s) 1st 100 0.6 1.2 2nd 250 0.9 1.4 Final Velocity of the Wooden Block: Vwbf = Xwb / t Final Velocity of the Pendulum Bob: Vbf = [2gL(1- cosθ)] ½ Angle: φ= h2 / L; θ = 90 0 – φ; h2 = L – h1 Moving Block Experiment A pendulum bob with length (L = 3m), vertically positioned is perpendicularly to a Wooden block of mass on a plane surface. The pendulum bob was raised to an angle (θ) to the horizontal and release to strike the wooden block for at least twice. See the table for details. As the bob hit the wooden block, it moved a distance of Xwb at a speed (m/s) as shown in the diagram. (Acceleration due to gravity = 9.8 m/s^2) Table 1 for the Data from Wooden Block Trial Mass (g) Xwb (meters) h1(meters) Initial Velocity(m/s) Final Velocity (m/s) 1st 100 0.6 1.2 2nd 250 0.9 1.4 Impulse of wooden block: Ft = ∆p (wooden block) = Mwb (Vwbf - Vwbi) Impulse of the ball: F∆t = ∆p (ball) = Mb (Vbf – Vbi) Momentum = Mass x velocity p = mv