Physics 1100: Collision & Momentum Solutions

1 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Physics 1100: Collision & Momentum Solutions1. The diagrams below are graphs of Force in kiloNewtons versus time in milliseconds for the motion of a 5 kg blockmoving to the right at m/s. (a) What is the magnitude and direction of the impulse acting on the block in each case? (b) What is the magnitude and direction of the average force acting on the block in each case? (c) What is the magnitude and direction of the final velocity of the block in each case?a. Impulse is given by the area under the F t curves. Since we have simple shapes, it is easy to find the area. Forrectangles area is height base and for triangles area is half the height I = 3 kN 3 ms = 9 N sii. I = 1 kN 6 ms = 6 N siii. I = 2 kN 2 ms + ( 4 kN) 2 ms = 0 N siv. I = (4 kN) 4 ms = 8 N sIf the impulse is positive, the net area was above the curve and it is directed to the right, if negative to the left.

2 B. We know I = Fave t where t is how long the Collision lasts. We read t from the graphs, so Fave = I/ Fave = (9 N s)/(3 ms) = 3000 Nii. Fave = ( 6 N s)/(6 ms) = 1000 Niii. Fave = (0 N s)/(4 ms) = 0 Niv. Fave = (8 N s)/(4 ms) = 2000 NIf the average force is positive it is directed to the right, if negative to the left. The impulse and force have the samedirection. c. Impulse is also equal to the difference in Momentum , I = mvf mvi. We can rearrange our equation for vf, vf =I/m + vf = (9 N s)/( kg) + 4 m/s = m/sii. vf = (?6 N s)/( kg) + 4 m/s = m/siii. vf = (0 N s)/( kg) + 4 m/s = 4 m/siv. vf = (8 N s)/( kg) + 4 m/s = m/s2. The diagrams below are the velocity versus time graphs for the Collision of motion of a 4 kg block with a wall. Thecollision lasts for 20 milliseconds in each case. (a) What is the magnitude and direction of the impulse acting on the block in each case?

3 (b) What is the magnitude and direction of the average force acting on the block in each case? a. Impulse is also equal to the difference in Momentum , I = mvf mvi. We have the mass, m = 4 I = (4 kg) ( 6 m/s 6 m/s) = 48 N sii. I = (4 kg) (2 m/s 8 m/s) = 24 N siii. I = (4 kg) (6 m/s 0 m/s) = +24 N sIf the impulse is positive it is directed to the right, if negative to the We know I = Fave t where t is how long the Collision lasts. We have already calculated I and we are given t =20 ms, so Fave = I/ Fave = ( 48 N s)/(20 ms) = 2400 Nii. Fave = ( 24 N s)/(20 ms) = 1200 Niii. Fave = (+24 N s)/(20 ms) = +1200 N3. You've been rowdy and obnoxious in a bar and now are in the process of being thrown out by the bouncer by the scruffof the neck. The bouncer has hold of you for s and you are given a final velocity of m/s. If your mass is kg,what was your final Momentum ?

4 What impulse and average force did the bouncer exert on you? Assume all motion is in astraight is defined by p = mv. Taking the direction of motion as positive, your initial Momentum was zero and yourfinal Momentum isp = ( kg)( m/s) = kg m/s .Impulse is defined as the change in momentumI = pf pi = kg m/s .Average force is related to impulse by I = Faveraget, soFaverage = I / t = kg m/s / 5 s = is the average force exerted on you and is in the same direction as your A ball of mass kg with speed m/s collides with a wall and bounces back with a speed of m/s. If themotion is in a straight line, calculate the initial and final momenta and the impulse. If the wall exerted a average force of1000 N on the ball, how long did the Collision last? Momentum is defined by p = mv. Taking the right as positive, the initial Momentum of the ball is pi = ( kg)( 15 m/s) = kg m/s.

5 The final Momentum ispf = ( kg)( m/s) = kg m/s .Impulse is defined as the change in momentumI = pf pi = kg m/s .Average force is related to impulse by I = Faveraget, and the wall would exert this force on the ball to the right. Therefore t = I / Faverage = kg m/s / +1000 = ball is in contact with the wall for approximately 13 A ball of mass kg glances of a wall as shown in the diagram. The ball approaches at 15 m/s at = 30 and leaves at12 m/s at = 20 . The Collision lasts for 15 milliseconds. (a) What are the components of the impulse experienced by the ball? (b) What are the components of the average force acting on the ball?a. We know Impulse is equal to the difference in Momentum , I = mvf vi. This is a vector equation and to getcomponents we consider the x and y components = mvfx mvix = ( ) (12cos20 15cos30 ) = N sIy = mvfy mviy = ( ) (12sin20 ( 15sin30 )) = + N sor I = + N We know I = Fave t where t is how long the Collision lasts.

6 We have already calculated I and we are given t =15 ms, so Fave = I/ tso Fave = ( + N s) / (15 ms) = + While chasing an armed suspect into and onto an ice rink, a police constable is shot. Fortunately, the constable is wearinga bullet proof vest which absorbs the bullet. If the muzzle velocity of the bullet is 350 m/s and the its mass is 100 g. Findthe final velocity of the constable and bullet if her mass is kg. Assume all motion is in a straight line and ignorefriction. Assume that the constable is at have a totally inelastic Collision , so Momentum is conserved. For this particular problem(mpolice + mbullet)vpf = mpolicevpi + we are told vpi = 0,vpf = mbulletvbi/(mpolice + mbullet) = ( kg)( 350 m/s)/( kg + kg) = m/s .So the constable is knocked backwards at m/s. 7. A 70 kg man and a 55 kg woman are standing on a stationary sled which is on a frictionless surface.

7 The man jumpshorizontally off the sled with a velocity of m/s at west of north. The woman jumps off the sled horizontally witha speed of m/s at south of west. What is the magnitude and direction of the sled's final Momentum ? If the massof the sled is kg, what is the final velocity of the sled? As is suggested by the word Momentum in this question, this is an explosion in which Momentum is conserved. Pman + Pwoman + Psled = 0 . (1) Momentum is a vector quantity so we will need to deal with the components. First we calculate the magnitude of themomentum of the man and the woman, using p = mv: Pman = 70 kg 3 m/s = 210 kg m/s ,Pwoman = 55 kg m/s = kg m/s .Examining equation (1), we see that Psled = (Pman + Pwoman), so we need to do a vector addition as shown in thediagram we find Pnet by components ijPman x= 210sin(25 ) = y= 210cos(25 ) = x= (40 ) Pwoman y= (40 ) = x= y= x= + y= the Pythagorean formula, Psled = [(Psled x)2+( Psled y)2] = [( )2+( )2] = kg m/s.

8 Using trigonometry, = arctan(Psled y/ Psled x) = arctan(| |) = .So the final Momentum of the sled is 238 kg m/s at south of find the final velocity of the sled recall that p = mv. This is a vector equation, so p and v must point in the samedirection. The magnitude of the velocity of the sled is thusvsled = Psled / msled = kg m/s / kg = m/s .So the velocity of the sled just after both people jump off the sled is m/s at south of A kg skater is travelling due east at a speed of m/s. A kg skater is moving due south at a speed of They collide and hold on to one another after the Collision , managing to move off at an angle south of east with aspeed vf. Find (a) the angle and (b) the speed vf, assuming that friction can be ignored. In any kind of Collision , Momentum is conserved so(m1 + m2)vf = m1v1i + m2v2i . (1)Now Momentum and velocity are vector quantities and the i and j components must be handled separately(m1 + m2)vfx = m1v1ix + m2v2ix , (1a)(m1 + m2)vfy = m1v1iy + m2v2iy.

9 (1b)So we can rearrange these equations to find the components of the final velocityvfx = (m1v1ix + m2v2ix) / (m1 + m2) , (2a)vfy = (m1v1iy + m2v2iy) / (m1 + m2) . (2b)Using the given values, we findvfx = [(50 kg)(3 m/s) + (70 kg)(0)] / (50 kg + 70kg) = m/s , (2a)vfy = [(50 kg)(0) + (70 kg)( 7 m/s) / (50 kg + 70 kg) = m/s . (2b)To find the magnitude and direction of the final velocity, we use the Pythagorean Theorem and trigonometry,vf = [(vfx)2 + (vfx)2] = m/s, and = arctan(|vfy/vfx|) = . The final velocity of the pair is m/s at south of Two opposing hockey players are racing up the ice for the puck when they collide at point A as shown in the diagrambelow. The first hockey player has mass 90 kg and a speed of m/s while the other has mass 82 kg and speed angle in the diagram is = 32 . After the Collision , the players remain locked together (at least until the referee forcesthem apart).]

10 What is the magnitude and direction of the players' velocity just after they collide?Since the Collision is totally inelastic and in two dimensions, we find that we are dealing with a vector addition problem, PT= P1 + P2. First we calculate the magnitude of each player's Momentum using p = mv,P1 = m1v1 = (90 kg)( m/s) = 243 kg m/s ,P2 = m2v2 = (82 kg)( m/s) = kg we find PT by the component method,ijP1x= 0P1y= 243P2x= (32 ) = (32 ) = the Pythagorean formula we find, PT = [(PTx)2 + (PTy)2] = [( )2 + ( )2] = kg m/s . Using trigonometry, we find the angle from = arctan(PTy/PTx) = arctan( ) = . So the total Momentum of the two players is PT = (478, ).Now PT = (m1 + m2)vf, so the final velocity must be in the same direction as the total Momentum . The magnitude of thevelocity isvf = PT/(m1 + m2) = kg m/s/ (90 kg + 82 kg) = m/s.