UNIT HOMEWORK MOMENTUM ANSWER KEY

1 UNIT HOMEWORK MOMENTUM ANSWER KEY MOMENTUM FORMULA & STUFF FROM THE PAST: p = m v, TKE = mv2, d = v t 1. An ostrich with a mass of 146 kg is running to the right with a velocity of 17 m/s. a. Calculate the MOMENTUM (in kg m/s) of the ostrich . Ans. p = m v = 146 kg 17 m/s = 2482 kg m/s b. Calculate the kinetic energy (in J) of the ostrich . Ans. TKE = mv2 = (146 kg)(17 m/s)2 = 21,097 J c. Calculate the distance (in m) travelled by the ostrich after 5 seconds. Ans. d = v t = 17 m/s 5 s = 85 m 2. A 21 kg child is riding a kg bike with a velocity of m/s to the left. a. Calculate the MOMENTUM (in kg m/s) of the child-bike system. Ans. p = m v = (21 kg + kg) m/s = ( kg) m/s = kg m/s b.

2 Calculate the kinetic energy (in J) of the child-bike system. Ans. TKE = mv2 = ( kg)( m/s)2 = J c. Calculate the distance (in m) travelled by the child-bike system after 5 seconds. Ans. d = v t = m/s 5 s = m COLLISIONS: ELASTIC 3. The ostrich from Probl. 1 collides with the child and his bike from Probl. 2. Assume the collision is elastic. Make the right the positive direction. a) What is the total MOMENTUM of the ostrich /child/bike system together (in kg m/s)? Ans. ptotal = postrich + pchild/bike = 2482 kg m/s kg m/s = kg m/s b) What is the total kinetic energy of the ostrich /child/bike system together (in J)? Ans. TKEtotal = TKEostrich + TKEchild/bike = 21,097 J + J = 21, J c) If the child-bike system bounces back with a velocity of 3 m/s, what is the final velocity of the ostrich (in m/s)?

3 Ans. m1v1 + m2v2 = m1v3 + m2v4 m2v4 m2v4 m1v1 + m2v2 m2v4 = m1v3 m1 m1 + = ( )( )+( . )( . ) ( . )( ) = . = . / = d) How much kinetic energy (in J) does the child/bike system have after the collision?

4 Ans. TKE = mv2 = ( kg)( 3 m/s)2 = J e) How much kinetic energy (in J) does the ostrich have? Ans. TKE = mv2 = (146 kg)( m/s)2 = 17, J f) Was kinetic energy conserved? YES NO TKEfinal TKEinitial. looks like the collision was not completely , J 21, J g) Was MOMENTUM conserved? YES NO kg m/s = kg m/s COLLISIONS: INELASTIC 4. The ostrich from Probl. 1 collides with the child and his bike from Probl. 2. Assume the collision is inelastic. a) What is the total MOMENTUM (in kg m/s) of the ostrich /child/bike system together? Ans. Nothing has changed. ptotal = postrich + pchild/bike = 2482 kg m/s kg m/s = kg m/s b) What is the mass (in kg) of the ostrich /child/bike system together?

5 Ans. 146 kg + kg = kg c) What is the final velocity (in m/s) of the ostrich /child/bike system together? Ans. m1v1 + m2v2 = (m1 + m2)v3 m1v1 + m2v2 = (m1 + m2)v3 (m1 + m2) (m1 + m2) + + = ( )( )+( . )( . )( + . )= .. =.

6 / = d) Will the system be moving to the right or to the left? Ans. The system will be moving in the positive direction which in this case is to the right. e) How much kinetic energy (in J) does the ostrich /child/bike system have? Ans. TKE = mv2 = ( kg)( m/s)2 = J f) Was kinetic energy conserved? YES NO Ans. TKEfinal TKEinitial. While total energy is conserved, some of the mechanical energy (kinetic) converted into work/heat. g) Was MOMENTUM conserved? YES NO Ans. Yes. pfinal = pinitial. kg m/s = kg m/s IMPULSE/ MOMENTUM : Impulse = F t, p = m v, d = v t 5. A kg football is thrown with a velocity of 15 m/s. A stationary receiver catches the ball and brings it to rest in s.

7 What is the force (in N) exerted on the receiver? Ans. F t = m v F= F=( . ) ( ) . = 375 N 6. A kg object is at rest. A N force acts on the object during a time interval of s. The force acts toward the right. a) What is the velocity (in m/s) of the object at the end of this interval? Ans. F t = m v = . ( . ) . = 9 m/s = b) At the end of this interval, a constant fore of N to the left is applied for s. What is the velocity (in m/s) at the end of the s? Ans. The force is to the left so it is negative.

8 We need to realize = and that we have an initial velocity, vi, of 9 m/s from the previous problem. F t = m v = = + + + = . ( . ) . +9 = 15 m/s = 7. A kg meteorite traveling at 30 m/s landed in Mario s front yard. It made a m hole in the ground. What force (in N) did it take to stop the meteorite? Ans. Ein = Eout TKEin = Wout mv2 = Fout dout mv2 dout=Fout ( kg)(30ms)2 ( m)=Fout 2250 N = Fout COLLISIONS TWO DIMENSIONS: p = m v 8.

9 A 2 kg object is at rest. A 4 kg object collides against it with a velocity of 5 m/s sending it flying off at an angle of 30 to the horizontal with a velocity of 8 m/s. What is the final velocity of the 4 kg object, if it moves off at an angle of 65 ? Ans. pin = pout m1 v1 = m1 va cos a + m2 vb cos b m2 vb cos b m2 vb cos b m1 v1 m2 vb cos b m1 v1 m2 vb cos b = m1 va cos a m1 v1 m2 vb m1 =va (4 kg)(5 ms) (2 kg)(8 ms) ( ) 4 kg ( )=va m/s = 9. A kg cue ball strikes the eight ball ( kg) elastically and the cue ball moves off at an angle of 45 to the horizontal at 3 m/s while the eight ball moves off at an angle of +60 to the horizontal at a velocity of 5 m/s.

10 What is the initial velocity (in m/s) of the cue ball? Ans. pin = pout m1 v1 = m1 va cos a + m2 vb cos b m1 v1 = m1 va cos a + m2 vb cos b m1 m1 v1=m1 va + m2 vb m1 v1=( kg)(3 ms) ( )+( kg)(5 ms) ( ) kg v1 = m/s BEFORE 4 kg 2 kg 5 m/s 2 kg 4 kg = 30 = 65 AFTER BEFORE kg kg ? m/s kg kg = 60 = 45 AFTER MOMENTUM : CONCEPTUAL QUESTIONS __(c)____ 1. A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped on the grass usually does not break because for the egg dropped on the grass, a) the change in MOMENTUM is less.