( ) Momentum and impulse Mixed exercise 1. 1 a. Using conservation of momentum: ( )

Transcription

1 Momentum and impulse Mixed exercise 1 1 a Using conseration of momentum: ( ) 6mu 4mu= 4m 1 u= After the collision the direction of Q is reersed and its speed is 1 u b Impulse = change in momentum I = (3m u) 0 = 6mu The magnitude of the impulse exerted by Q on P is 6mu a = u + as = = 14ms The speed of the drier immediately before it hits the pile is 14 m s 1 b Using conseration of momentum: ( ) = 100 = 35 3 The common speed of the pile and pile drier is 35 3 m s 1 c First, use F = ma to find the deceleration. 100g = 100a a= 90. u as ( ) = = s 3 s= 0.75m( s.f.) Use the common speed found in part b for u. s.f. as g = 9.8 has been used. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 d The model assumes that the pile drier would not bounce upon contact with the pile, i.e. the particles coalesce. Gien that the pile drier is much heaier than the pile, this would be a fair assumption. 3 a u = 18, = 1, t =.4, a =? = u+ at 1= 18+.4a 1 18 a= =.5.4 F = ma F = 800 (.5) = 000 F = 000 You are going to hae to use F = ma to find F. So the first step of your solution must be to find a. The retarding force is slowing the car down and is in the negatie direction. So, in the positie direction, the force is F. The alue of F is 000. b u= 18, = 1, t=.4, s=? u+ s= t =.4= 15.4= 36 The distance moed by the car is 36 m You could use the alue of a you found in part a and another formula. Unless it causes you extra work, it is safer to use the data in the question. 4 a Conseration of momentum 0. 4 = (0. ) + ( ) 0.8= = = A full formula for the conseration of momentum is m A u A + m B u B = m A A + m B B. In this case the elocity of B is zero. The speed of A after the impact is 1.75 m s 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.

3 4 b Consider the impulse of B on A I = m mu = ( ) (0. 4) = = 0.45 The magnitude of the impulse of B on A during the impact is 0.45 N s It is a common mistake to mix up the particles. The impulses on the two particles are equal and opposite. Finding the magnitude of the impulse, you can consider either particle either would gie the same magnitude. Howeer, you must work on only one single particle. Here you can work on A or B, but not both. 5 a Conseration of linear momentum = (000 ) + (3000 5) 0 000= = = The speed of P immediately after the collision is.5 m s 1 b For Q, I = m mu I = (3000 5) (3000 0) = To find the magnitude of the impulse you could consider either the change in momentum of P or the change of momentum of Q. You must not mix them up. The magnitude of the impulse of P on Q is N s 6 You do not know which direction Q will be moing in after the impact. Mark the unknown elocity as m s 1 in the positie direction. After you hae worked out, the sign of will tell you the direction Q is moing in. a Conseration of momentum (1.5 3) + (.5 ( 4)) = (1.5 (.5)) + (.5 ) = = = = = The sign of is negatie, so Q is moing in the negatie direction. It was moing in the negatie direction before the impact and so its direction has not changed. The speed of Q immediately after the impact is 0.7 m s 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 6 b The direction of Q is unchanged. c For P, I = m mu I = (1.5 (.5)) (1.5 3) = 8.5 The magnitude of the impulse exerted by Q on P is 8.5 N s 7 After the collision A (of mass m) and B (of mass km) combine to form a single particle. That particle will hae the mass which is the sum of the two indiidual masses, m + km. Conseration of momentum: ( ) ( m u) + ( km ( u)) = ( m+ km) u 3 mu kmu= mu+ kmu 3 3 mu mu= kmu+ kmu mu= kmu k = = The total linear momentum before impact must equal the total linear momentum after impact. Particle B is moing in the negatie direction before the collision and so it has a negatie linear momentum. m and u are common factors on both sides of the equation and can be cancelled. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 8 a After impact, the sledgehammer and the metal pin moe together. You model the sledgehammer and pin as a single particle of mass 1 kg. b Conseration of momentum: ( ) (10 9) + ( 0) = 1 90 = = The speed of the pin immediately after impact is 7.5 m s 1 The model gien in the question assumes that the pin and sledgehammer stay in contact and moe together after impact, before coming to rest. Although the question only refers to the pin, you must consider the pin and the sledgehammer as moing together, with the same elocity and the same acceleration, throughout the motion after the impact. u = 7.5, = 0, s = 0.03, a =? = u + as 0 = ( a 0.03) 7.5 a= = Using F = ma: 1g R= 1 ( 937.5) R= (1 9.8) + ( ) = The alue of R is ( s.f.) c The resistance (R) could be modelled as arying with speed. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 9 Impulse = change in momentum = 0.5(3i+ 0 j) 0.5( 5 i) = (4i+ 10 j)ns Magnitude of the impulse= Ns = 6 Ns Angle between the impulse and the direction i is α where 10 tanα = 4 α = 3 (nearestdegree) 10 Let elocity before being hit be u m s 1 impulse = change in momentum.4i+ 3.6j= 0.(1i+ 5 j) 0.u 0.u=.4i+ j.4i 3.6j =.6j u= 13j The elocity of the ball immediately before it is hit is 13j m s 1 11 Let the elocity of Q after the collision be m s 1 Use conseration of momentum: 4(i+ 16 j) + 3( i 8 j) = 4( 4i 3 j) + 3 5i+ 40j= 16i 18j+ 3 3= 1i+ 168j = 7i+ 56j The elocity of Q immediately after the collision is (7i + 56j) m s 1 1 a 3 r= ( t + t + 4 t) i+ (11 t) j = rɺ = (3t + t+ 4) i+ 11j When t = 4, = (60) i+ 11j Differentiate displacement ector to gie the elocity ector. = = 61 The speed of P when t= 4 is 61 m s 1 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 1 b Let the elocity immediately after the impulse be V m s 1 Then as impulse = change in momentum.4i+ 3.6j= 0.3V 0.3(60i+ 11 j) 0.3V=.4i+ 3.6j+ 18i+ 3.3j = 0.4i+ 6.9j V= 68i+ 3j Use impulse = change in momentum. elocity of P immediately after the impulse is (68i + 3j) m s 1 Challenge Using equations for impulse 1 Q changes direction after impact: u km( + u) = m(u ) so k= u + P changes direction after impact u+ km(u ) = m( + u) so k= u a k must be positie so u > u b If k= then Q changes direction after impact. u + u+ If k= then P changes direction after impact. u Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7