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Ch 8 HW Day 5 Solus: 78, 82, 86, 87 *78 •• Picture the Problem We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the colliding objects that we can solve for v1f and v2f. Apply conservation of momentum to the elastic collision of the particles to obtain: m1v1f m2v2f m1v1i m2v2i (1) m1v12f 12 m2 v22f 12 m1v12i 12 m2 v22i Relate the initial and final kinetic energies of the particles in an elastic collision: 1 2 Rearrange this equation and factor to obtain: m2 v22f v22i m1 v12i v12f or m2 v2f v2i v2f v2i m1 v1i v1f v1i v1f Rearrange equation (1) to obtain: Divide equation (2) by equation (3) to obtain: (2) m2 v2f v2i m1 v1i v1f v2f v2i v1i v1f (3) Rearrange this equation to obtain equation (4): v1f v2f v2i v1i (4) Multiply equation (4) by m2 and add it to equation (1) to obtain: Solve for v1f to obtain: Multiply equation (4) by m1 and subtract it from equation (1) to obtain: Solve for v2f to obtain: m1 m2 v1f m1 m2 v1i 2m2v2i v1 f m1 m2 2m2 v1i v2 i m1 m2 m1 m2 m1 m2 v2f m2 m1 v2i 2m1v1i 2m1 m2 m1 v2 f v1i v 2i m1 m2 m1 m2 Remarks: Note that the velocities satisfy the condition that v2f v1f v2i v1i . This verifies that the speed of recession equals the speed of approach. 82 • Picture the Problem Let the mass of the bullet be m, that of the wooden block M, the pre-collision velocity of the bullet v, and the post-collision velocity of the block+bullet be V. We can use conservation of momentum to find the velocity of the block with the bullet imbedded in it just after their perfectly inelastic collision. We can use Newton’s 2nd law to find the acceleration of the sliding block and a constant-acceleration equation to find the distance the block slides. Using a constantacceleration equation, relate the velocity of the block+bullet just after their collision to their acceleration and displacement before stopping: Solve for the distance the block slides before coming to rest: Use conservation of momentum to relate the pre-collision velocity of the bullet to the post-collision velocity of the block+bullet: 0 V 2 2ax because the final velocity of the block+bullet is zero. V2 x 2a mv m M V (1) Solve for V: m V v mM Substitute in equation (1) to obtain: 1 m x v 2a m M Apply F ma to the block+bullet (see the FBD in the diagram): F x (3) and F y Use the definition of the coefficient of kinetic friction and equation (4) to obtain: 2 (2) f k m M a Fn m M g 0 f k k Fn k m M g Substitute in equation (3): k m M g m M a Solve for a to obtain: a k g Substitute in equation (2) to obtain: 1 m x v 2 k g m M 2 (4) Substitute numerical values and evaluate x: 1 x 20.22 9.81 m/s 2 2 0.0105 kg 750 m/s 0.130 m 0.0105 kg 10.5 kg 86 •• Picture the Problem This nuclear reaction is 5Li + p + 3.15 1013 J. To conserve momentum, the alpha particle and proton must move in opposite directions. We’ll apply both conservation of energy and conservation of momentum to find the speeds of the proton and alpha particle. Use conservation of momentum in this process to express the alpha particle’s velocity in terms of the proton’s: pi pf 0 and 0 mp vp m v mp mp Solve for v and substitute for m to obtain: v Letting E represent the energy released in the reaction, apply conservation of energy to the process: K p K E Substitute for v: m vp 4mp vp 14 vp or 1 2 mp vp2 12 m v2 E 1 2 mp vp2 12 m 14 vp E 2 Mult by 32 (least common denominator) to get rid of all fractions! Solve for vp and substitute for m to obtain: Substitute numerical values and evaluate vp: vp vp 32 E 32 E 16mp m 16mp 4mp 32 3.15 10 13 J 20 1.67 10 27 kg 1.74 10 7 m/s Use the relationship between vp and v to obtain v: 87 v 14 vp 14 1.74 107 m/s 4.34 106 m/s ••• Picture the Problem The pictorial representation shows the projectile at its maximum elevation and is moving horizontally. It also shows the two fragments resulting from the explosion. We chose the system to include the projectile and the earth so that no external forces act to change the momentum of the system during the explosion. With this choice of system we can also use conservation of energy to determine the elevation of the projectile when it explodes. We’ll also find it useful to use constant-acceleration equations in our description of the motion of the projectile and its fragments. (a) Use conservation of momentum to relate the velocity of the projectile before its explosion to the velocities of its two parts after the explosion: The only way this equality can hold is if: pi pf m3 v 3 m1v1 m2 v 2 m3v3ˆi m1vx1ˆi m1v y1ˆj m2v y 2ˆj m3 v3 m1v x1 and m1v y1 m2 v y 2 Express v3 in terms of v0 and substitute for the masses to obtain: vx1 3v3 3v0 cos 3120 m/s cos30 312 m/s and v y1 2 v y 2 (1) Using a constantacceleration equation with the downward direction positive, relate vy2 to the time it takes the 2kg fragment to hit the ground: With Ug = 0 at the launch site, apply conservation of energy to the climb of the projectile to its maximum elevation: Solve for y: Substitute numerical values and evaluate y: Substitute in equation (2) and evaluate vy2: Substitute in equation (1) and y v y 2 t g t 2 1 2 y 12 g t t 2 vy2 (2) K U 0 Because Kf = Ui = 0, or Ki U f 0 12 m3v y20 m3 gy 0 y v y20 2g 2 v0 sin 30 2g 120 m/s sin30 2 y 2 9.81m/s 2 vy2 183.5 m 183.5 m 12 9.81m/s 2 3.6 s 3.6 s 33.3 m/s v y1 233.3 m/s 66.6 m/s 2 evaluate vy1: Express v in vector form: 1 v1 vx1iˆ v y1 ˆj (b) Express the total distance d traveled by the 1kg fragment: Relate x to v0 and the time-toexplosion: Using a constantacceleration equation, express texp: Substitute numerical values and evaluate texp: Substitute in equation (4) and evaluate x: Relate the distance traveled by the 1kg fragment after the explosion to the time it takes it to reach the ground: 312 m/s iˆ 66.6 m/s ˆj d x x' (3) x v0 cos t exp (4) texp v0 sin g g texp vy0 120 m/s sin30 6.12 s 9.81 m/s 2 x 120 m/s cos306.12 s 636.5 m x' vx1t' Using a constantacceleration equation, relate the time t for the 1kg fragment to reach the ground to its initial speed in the y direction and the distance to the ground: y v y1t' g t' 1 2 2 183.5 m = 66.6 m/s t – 4.905 t2 Use QF, Polysmelt, or graph Substitute to obtain the quadratic equation: Solve the quadratic equation to find t: t = 15.9 s Substitute in equation (3) and evaluate d: d x x' x vx1t' 636.5 m 312 m/s 15.9 s 5.61km (c) Express the energy released in the explosion: Eexp K K f K i (5) Find the kinetic energy of the fragments after the explosion: K f K1 K 2 12 m1v12 12 m2v22 1kg 312 m/s 2 66.6 m/s 2 2 12 2 kg 33.3 m/s 1 2 52.0 kJ Find the kinetic energy of the projectile before the explosion: K i 12 m3v32 12 m3 v0 cos Substitute in equation (5) to determine the energy released in the explosion: Eexp K f Ki 52.0 kJ 16.2 kJ 2 1 2 3 kg 120 m/s cos 302 16.2 kJ 35.8 kJ