Answers to Dynamics Homework

Force Acceleration and Motion (Pages 1-3)

   1.   B		 (1992 PI  Q7)
   2.   B		 (1998 PI  Q5)
   3.   B		 (1996 PI  Q6)
   4.   C	 	 (1994 PI  Q5)
   5.   A		 (1995 PI  Q5)
   6.   B		 (1995 PI  Q6)
   7.   C		 (1994 PI  Q4)
   8.   E		 (1996 PI  Q7)

 9.a. 			(1998 PII Q2a/b/c/d)
   b.  a = 1.5m/s2
   c.  a = -1.5m/s2
       This can be thought of as an upward deceleration of 1.5m/s2
   d. 


10.i.	a = 2.7m/s/s	(1997 PII Q1b)
  ii.   s = 201.67m



Slope Problems (Pages 4-6)

  11.      C 			(1993 PI Q3)  
  12.      A			(1994 PI Q6)   
  13.      Fresultant = 5.8N	(1995 PI Q33)    
  14.a.i.  Fslope = 136.8N 	(1996 PII Q2ai/ii)
     a.ii. a = 3.35m/s2
     b.				(1996 PII Q2b)			
     c.				(1996 PII Q2c)
   
  15.a.i.    Fs = 6.7N		(1998 PII Q1ai/ii)
     a.ii.
			
     a.iii.				(1998 PII Q1aiii/iv/v)		
     a.iv. t = 0.75s			
     a.v.  s = 1.125m
     b.i.				(1998 PII Q1bi)
		
     b.ii.				(1998 PII Q1bii)

     
  16.a. 1.5s				(1992 PII  Q1a/b/c)   
     b. 

     c.

     d.			(1992 PII  Q1d)

Momentum (Pages 7-14)

 17.   A 			(1993 PI  Q5) 
 18.   D			(1994 PI  Q8)
 19.   C			(1996 PI  Q8)
 20.   E			(1992 PI  Q9)
 
 21.a. 	20kg/ms-1
    b. 	500kg/ms-1(downwards)
    c. 	3750kg/ms-1
    d. -3kg/ms-1(-ve indicates direction)

 22.	20ms-1
 23.   -1.5ms-1
 24.	46,666.67kg (sum of both parts)
 25.   -7.5m/s	
 26.    50kg

 27.a.  90,000J
    b.	4325J
    c.  6.25x109J 
 
 28. 	Ek(before) = 11J
	Ek(after) =  8.5J
 	Ek not conserved => inelastic collision

 29.	Ekbus= 500,000J
	Ekcar= 1,125,000J...Largest kinetic energy.

	P(bus)= 100,000kgms-1...Largest momentum.
	P(car)= 45,000kgms-1

 30.i.	v = 8000ms-1
   ii.	v = 565.7ms-1

 31.	v = -2.67ms-1
	
 32.a.	A rocket expels hot gases at a rate of 
	many kilogrammes per second. To conserve
	the total momentum the rocket must be 
	propelled in the opposite direction to 
	the hot gases. This example is like an
	explosion.

    b.	The explanation for the aeroplane is the 
	same as that of the rocket. The only difference
	is the rate that the gases are expelled
	from the engines.
	
 33.	   uball = 12.00m/s					(1998 PI Q32)
 34.a.	   Pbefore = Pafter(no external forces affecting motion)	(1993 PII Q3)
    b.i.   uA = 20m/s
    b.ii.  Ffriction = 13500N
    b.iii. Most of the kinetic energy will be converted into heat energy.

 35.a.i.   v = 1.4m/s						(1995 PII Q3ai/aii)
    a.ii.  ubullet = 560m/s
    
    b. 								(1995 PII Q3b)
   
 36.a.								(1992 PII Q3a)
    bi.	 Pbefore = Pafter(no external forces affecting motion)	(1992 PII Q3bi/bii)
    bii. upellet = 100.5m/s
    c.								(1992 PII Q3c)

 37.a.   Pbefore = Pafter(no external forces affecting motion)	(1999 PII Q3a)
    b.i. 							(1999 PII Q3bi)
    b.ii.							(1999 PII Q3bii)

 38.a.	Collision with polyurethane block.			(1994 PII Q4a)
	Ekafter = 0.027J
	
	Collision with rubber band.
	Ekafter = 0.046J

   b.								(1994 PII Q4b)
   c.								(1994 PII Q4c)
	

Impulse(Pages 15-21)

 39.    C		(1999 PI  Q7)
 40.    D		(1994 PI  Q7)
 41.    C		(1997 PI  Q9)
 42.    C		(1996 PI  Q9)
 43.    A		(1997 PI  Q8)
 44.    B		(1993 PI  Q7)
 
 45.a.	Favg = 1.67N 	(1992 PI  Q32a)
    b.	
      	
 46.	  Favg = 3200N
 47.	  m = 8kg
	
 48.a.i.  V = 0.4m/s			   (1998 PII  Q3ai)
    a.ii. Favg = 3.6N	   		   (1998 PII  Q3aii)
    b.i.  Contact time uncertainty = 20%   (1998 PII  Q3bi)  
    b.ii. Favg = (3.6 +- 0.72)N 		   (1998 PII  Q3bii)
	
 49.a.					   (1997 PII  Q3a)
    b.	  DPp = -0.12kgm/s		   (1997 PII  Q3b)
    c.i.  Faverage  = 2N			   (1997 PII  Q3ci)
    c.ii.

 50.a.	  VA = 0.3m/s
    b.    Ek(before) = 0.09J
	  Ek(after) =  0.05J
 	  Ek not conserved => inelastic collision
    c.	  Favg = -7.5N

 51.i.    Faverage  = -5000N  		(1996 PII  Q3bi)
    ii.   P = 1x107Pa			(1996 PII  Q3bii)
    iii.				(1996 PII  Q3biii)

 52.a.	  t = 3.6s			(1999 PII  Q2a)
    b.    Fun = 623.3N			(1999 PII  Q2b)
    c.					(1999 PII  Q2c)

 53.a.	   a = -9.0m/s/s		(1993 PII  Q1a)
    b.	  			   	(1993 PII  Q1b)
    c.	   DP = 6.8kgm/s		(1993 PII  Q1c)
    d.     F = 13.6N			(1993 PII  Q1d)
 

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