Chapter 9; Linear Momentum and Collisions
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D9.1 What is the momentum of a 1200 kg sedan traveling at 90 km/hr? At what speed must a 3600 kg truck travel to have the same momentum? First, change the speed to units of m/s, v = 90 = 25p = m v = [1200 kg] = 30 000 ptruck = ptruck vtruck vtruck = = = 8.33 vtruck = 8.33 = 30 D9.2 A 150 gram baseball initially traveling at 30 m/s is struck by a bat and leaves in the opposite direction at 35 m/s. a) What is its change in momentum? b) Is the change in momentum directed along the initial velocity, the final velocity, or some other direction? c) What is the impulse delivered by the bat? After the dive, the total momentum is still zero:
PTot, final = pLeftDiver + praft + pRightDiver v = - 2 m/s That is, the raft moves to the left at 2 m/s. Initially, PTot, initial = (250 g)(25 cm/s) = 6250 gm cm/s.After the collision, PTot, final = (850 g) v = 6250 gm cm/s = PTot, initial D9.8 A 25 g bullet is fired horizontally with a speed of 100 m/s into a 5.0 kg block of wood; the wooden block and bullet start to move off with speed vf. What is this speed vf? The block is suspended by cords so that it moves in an arc (but does not rotate). What is the height h to which it rises? PTot, final = (5.025 kg) vf = (0.025 kg)(100 m/s) = PTot, initialvf = 0.50 m/s As the block with embedded bullet starts to move, it has KE given by KE = (1/2) M v2 = (1/2) (5.025 kg) (0.50 m/s)2 = 0.62 JRelative to the bottom of its swing, its PE is now zero So its total energy is E = KE + PE = 0.62 J At the top of its swing, the block (with embedded bullet) momentarily comes to rest so its KE there is zero and all of its energy is now PE = m g h E = KE + PE = 0 + m g h = m g hBy energy conservation, this final energy must still be 0.62 J E = m g h = (5.025 kg)(9.8 m/s2) h = 0.62 Jh = 0.0126 m = 1.26 cm D9.9 A stationary 238U nucleus with atomic mass 238 x 1.67 x 10-27 kg decays by emitting an alpha particle with atomic mass 4 x 1.67 x 10-27 kg at a speed of 1.5 x 107 m/s. What will be the recoil velocity of the resulting 234Th nucleus? The "pattern" should be clear by now. Initially, PTot, initial = 0After the radioactive decay, the alpha carries momentum to the right and the thorium nucleus recoils and carries momentum to the left. The total momentum is still zero. PTot, final = (234 m) (- v) + (4 m) (1.5 x 107 m/s) = 0 = PTot, initialwhere m = 1.67 x 10-27 kg v = 2.56 x 105 m/s D9.9 A 600 g glider moves on an air track with a velocity of 10 cm/s to the right while a 400 g glider moves to the left with a velocity of 20 cm/s. The two collide elastically. What is the velocity of each after the collison? Momentum is always conserved, PTot,fin = (600 g) v1f + (400 g) v2f = (600 g)(10 ) + (400 g)(- 20 ) = PTot,in(600)v1f + (400)v2f = - 2000 3v1f + 2v2f = - 10 or, if we drop the units, this becomes 3v1f + 2v2f = - 10This is one equation with two unknows, so we must find additional information. Since this is an elastic collision, we also know that the Kinetic Energy is conserved. = (1/2)(600 g)(10 cm/s)2 + (1/2)(400 g)(- 20 cm/s)2 = KETot,in 3 v1f 2 + 2 v2f2 = 3 (10 )2 + 2( - 20 )2 3 v1f 2 + 2 v2f2 = 1100 Now we have two equations in two unknowns. We can solve for v2f from the earlier one and substitute it into this equation, 3v1f + 2v2f = - 102v2f = - 3v1f - 10 v2f = - 1.5v1f - 5 3 v1f 2 + 2 [- 1.5v1f - 5]2 = 1100 3 v1f 2 + 2 [2.25v1f2 + 15 v1f + 25] = 1100 3 v1f 2 + 4.5v1f2 + 30 v1f + 50 = 1100 7.5v1f2 + 30 v1f - 1050 = 0 There are two possible solutions, due to the ± in the equation;
That is, either v1f(1) = - 14 and v1f(2) = 10 or The second set is somewhat interesting for D9.10 Below are eight sets of initial conditions for perfectly elastic collisions on an air track. Glider number two is at rest in each case. Calculate the velocity of glider number two in each case and rank them in order of increasing velocity of glider number one.
The text gives the velocity v2f for just such a case, . Therefore, = = = 1,633
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