Chapter 9. particle is increased.


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1 Chapter 9 9. Figure 936 shows a three particle system. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the three particle system. (c) What happens to the center of mass as the mass of the m 3 particle is increased. We can calculate the center of mass in the x and y directions. x cm (m + m 2 + m 3 ) (m x + m 2 x 2 + m 3 x 3 ) (3.0kg 0m + 4.0kg 2.0m + 8.0kg.0m) (3.0kg + 4.0kg + 8.0kg) (5kg) (6.0kgm) 6 5 m y cm (m + m 2 + m 3 ) (m y + m y + m y ) (3.0kg 0m + 4.0kg.0m + 8.0kg 2.0m) (3.0kg + 4.0kg + 8.0kg) (5kg) (20.0kgm) 20 5 m 4 3 m As the topmost mass grows larger, the cm will move toward that mass.
2 9.5 In the ammonia (NH3) molecule of Fig 939, three hydrogen atoms form an equilateral triangle, with the center of the triangle at a distance d m from each hydrogen atom. The nitrogen atom (N) is at the apex of a pyramid, with the three hydrogen atoms forming the base. The nitrogen to hydrogen atomic ratio is 3.9 and the nitrogen to hydrogen distance is L m. What are (a) x and (b) y coordinates of the molecules center of mass. (a). y symmetry, we can see that the x coordinate of the center of mass is 0. (b) To find the y component, we replace the 3 H atoms, with one particle with the mass of the three H atoms at the cm of the three H. The cm of the 3H is at (0,0) by symmetry. We now find the y CM by for the N and the 3H particle y N L 2 d 2 (0.4 0 ) 2 (9.4 0 ) m y 3H 0.0m m N 3.9m H y 3m H y 3H + m N y N 3m H + m N 3m H m H m 3m H +3.9m H m m Two skaters, one with mass 65 kg and the other with mass 40 kg stand on an ice rink hold a ole with a length of 0m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far will the 40 kg skater move? Assume that the 65 kg is at 0 and the 40 kg is at 0m. The cm does not move, so both skaters end up at the cm. The 40 kg skater moves from 0 to the cm. x cm m The 40 kg skater moves 6.9 m.
3 9.5 A shell is fired from a gun with a muzzle velocity of 20m/s at an angle of 60. At the top of the trajectory, the shell explodes into two equal mass fragments. One fragment, whose speed immediately after the explosion is zero falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible. Let s sketch what happens first Examine how long it took to reach the explosion point. v fy 0 v iy 20m / s sin m / s v ix 20m / s cos60 0m / s a g v fy v iy gt t v iy g.767s Next we find the distance traveled before the explosion. x L v ix t 7.67m The left halfshell lands at x L 7.67m. If the shell had not exploded, we know that it would land at x cm m 35.34m Since the forces involved in the explosion were entirely internal, the center of mass of the two shells still lands at exactly this point. Knowing where the cm is allows us to find where the second half shell will land.
4 x cm 2 m x + L 2 m x R m x cm 2 x L + 2 x R x R 2x cm x L m 9.8 A 0.70 kg ball is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in the linear momentum of the ball Δ p 0.7kg 5m / s 0.7kg ( 2 m / s) 4.9kgm / s 9.26 A.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 0 m/s. It rebounds with an initial speed of 0 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for s, what is the magnitude of the average force on the floor from the ball Δ p.2kg ( 25m / s).2kg (0m / s) 42kg m / s F Δ p Δ t 42kg m / s 0.020s 200N 9.35 A 9 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of 4 m/s. What speed does he acquire as a result? This is a momentum conservation problem. The total momentum is zero (man and stone at rest at the beginning). The total momentum must remain zero. In the final state
5 0 p m + p s p m p s m m v m m s v s v m m s v m s m 0.068kg 9.0kg 4m / s m / s 9.42 In Fig 957, a stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction µ L 0.40 and slides to a ston a distance d L 0.5m. Piece R encounters a coefficient of kinetic friction µ R 0.50and slides to a ston distance d R 0.25m What was the mass of the block? Since we know the mass of the L block, we ll begin with it. The L block stops because friction does work on it. We can compute the work done W L F f d L µ L N L d L µ L m L gd L kg 9.8m / s 2 0.5m.76J Now that we know the work done, we can compute the velocity it had after the explosion. This was the velocity it had when it hit the friction K f K i W.76J 0 2 m L v L 2 W v L 2W 2 (.76J) m L 2kg.08m / s Now lets consider the right side. As before, we can compute the velocity from the work done.
6 K f K i W 0 2 m R v R 2 W 2 m v 2 R R µ R m R gd R v R 2µ R g d R.565m / s Now we know the velocities. Given that the momenta of the right and left pieces must be equal in magnitude, we can solve for the unknown mass. m L v L m R v R m R v L v R m L m R.08m / s.565m / s 2kg.38kg 9.45 A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden block at rest on a frictionless surface. The bullet emerges traveling in the same direction with its speed reduced to 428 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bulletblock center of mass? This is a conservation of momentum problem. m b v b + M v kg 672m / s + 0.7kg 0m / s kgm / s m bv b + M v kg 428m / s + 0.7kg v kg m / s m b v b + M v kg 428m / s + 0.7kg v v kg m / s kg 428m / s 0.7kg.82m / s We can find the CM velocity because it remains unchanged (since the forces are all internal to the bulletblock system) v cm m bv b + M v m b + M kg m / s kg + 0.7kg 4.955m / s 9.46 A bullet of mass 0 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 2 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet s initial speed.
7 This is both a momentum conservation problem and an energy conservation problem. The energy conservation part is the motion of the pendulum. The total energy of the pendulum at the top of the swing is equal to the total energy of the pendulum after it has been struck by a bullet. E f (M + m)g y E i 2 (M + m)v 2 E i E f 2 (M + m)v 2 (M + m)g y v 2g y m / s Now that we know the velocity of the bulletpendulum immediately after the bullet strikes the pendulum, we can use momentum conservation to find the bullet s velocity mv b (M + m)v (M + m) v b v m (2.0kg + 0.0kg).534m / s 0.0kg 308.3m / s 9.50 In Fig. 960, a 0 g bullet moving directly upward at 000m/s strikes nd passes through the center of mass of a 5.00 kg block initially at res. The bullet emerges from he block moving directly upward at 400 m/s. Tow what maximum height does the block then rise above its initial position. As in the ballistic pendulum case, this problem uses both momentum conservation and energy conservation. We begin with momentum conservation in the collision. m b v b + M v kg 000m / s kg 0m / s 0 kgm / s kg 400m / s + 2.0kg v 0.0 kg m / s m bv b + M v m b v b + M v kg 400m / s kg v v 0.0 kg m / s kg 400m / s 2.0kg 3m / s
8 Now we know the velocity of the block after the collision. We can use conservation of energy to find the maximum height. 2 Mv2 Mgh h v 2 (3m / s)2 2g 2 9.8m / s m 9.60 A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long an fixed at the far end. The ball is then released when the cord is horizontal (Fig. 965). At the bottom of its path, the ball strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision. We begin by finding the initial velocity of the ball using energy conservation. 2 mv 2 mgh v 2 2gh v b 2gh m 3.7m / s Now consider the collision. oth momentum and energy are conserved in this elastic collision. mv b m v b + M v 2 mv 2 b 2 m v 2 b + 2 M v 2 We have two equations and two unknowns. We solve the momentum equation for one of the unknowns and then plug into the energy equation.
9 m v b mv b M v v b v b M v m 2 mv 2 b 2 m v 2 b + 2 M v 2 2 mv 2 b 2 m(v M b m v )2 + 2 M v 2 mv 2 b 2 mv 2 b + 2 M 2 m v 2 Mv b 0 M 2 v m 2 2 2Mv b v + M v 0 M m v 2 2 2v b v + v M v m + v 2v b M 0 m + v 2v b 2v v b 2 3.7m / s M m + 2.5kg 0.5kg +.23m / s 2 v + 2 M Notice that if the mass of the ball and the mass of the block are the same, the ball will stop and the block will take all of the velocity of the ball. This is exactly what we saw in the elastic equal mass collision in lab. We finally find the ball s velocity after the collision, by substituting back in. v b v b M v m 3.7m / s m / s 2.45m / s v 2
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