Circle correct course: PHYS 1P21 or PHYS 1P91 BROCK UNIVERSITY. Course: PHYS 1P21/1P91 Number of students: 260 Examination date: 10 November 2014

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1 Tutorial #: Circle correct course: PHYS P or PHYS P9 Name: Student #: BROCK UNIVERSITY Test 5: November 04 Number of pages: 5 + formula sheet Course: PHYS P/P9 Number of students: 0 Examination date: 0 November 04 Time of Examination: 3:00 3:50 Instructor: S. D Agostino No aids are permitted except for a non-programmable, non-graphics calculator. Solve all problems in the space provided. Total number of marks: 0. [ point] Ball A has mass 0 kg and Ball B has mass kg. The two balls are at rest 5 m apart. The centre of mass of the two balls is (a) closer to Ball A. (b) closer to Ball B. (c) half-way between the two balls. (d) [The centre of mass makes sense for only a single object, not two objects.]. [ point] The net external force on a system is not zero. (a) The energy and momentum of the system are conserved. (b) The energy of the system is conserved but the momentum of the system is not conserved. (c) The momentum of the system is conserved but the energy of the system is not conserved. (d) Neither the energy nor the momentum of the system are conserved. (e) [Not enough information is given.] 3. [ point] A person carries a book while walking in a straight line at a constant speed on level ground. The book s distance from the ground remains constant. The work done by the person on the book is (a) positive. (b) negative. (c) zero. (d) [It depends on which is the positive direction.] (e) [It depends on the relative magnitudes of the weight of the book and the normal force exerted by the person on the book.]

2 4. [ point] Alice throws a ball to her friend Basil. As Basil catches the ball, the work he does on the ball is (a) positive. (b) negative. (c) zero. (d) [It depends on whether the direction of her throw is the positive direction or the negative direction.] 5. [4 points] Block moves to the right with speed.7 m/s and Block moves to the left with speed 4.8 m/s. The two blocks collide head-on and stick together. The mass of Block is three times the mass of Block. Determine the velocity of the blocks just after the collision. Solution: Draw a diagram! I choose to the right as the positive direction. Let v represent the velocity of the blocks just after they stick together, and let v and v respectively represent the velocities of the two blocks before the collision. Then by the principle of conservation of momentum, (m + m ) v = m v + m v v = m v + m v m + m Now use the fact that m = 3m, to obtain v = m v + 3m v m + 3m v = m (v + 3v ) 4m v = v + 3v ( 4.8) v = 4 v =.93 m/s The speed of the two blocks just after the collision is.93 m/s, and they move to the left.. [4 points] A block of mass 0.4 kg slides up a ramp. The block has an initial speed of.4 m/s and friction in the ramp causes 5.8 J of the block s mechanical energy to be converted to thermal energy by the time it stops. What is the vertical component of the block s displacement by the time it stops? Solution: Draw a diagram! Use i to label the initial position of the block and use f to label the final position of the block. Let E represent the amount of mechanical

3 energy that is transformed into thermal energy by the end of the motion. Then by the principle of conservation of energy, U f + K f + E = U i + K i U f U i = K i K f E mgy f mgy i = mv i mv f E mg [y f y i ] = mv i mv f E mg y = mv i mv f E mg y = mv i m(0) E mg y = mv i E y = mv i E mg y = (0.4)(.4) (5.8) (0.4)(9.8) y = 0. m Thus, as the block slides up the ramp its vertical component of displacement is cm. 7. [4 points] A pendulum consists of a block of mass 0.9 kg hanging from one end of a string. m long. The other end of the string is attached to the ceiling. When the block is at rest at its lowest point it is given a horizontal velocity v. The block rises and eventually stops momentarily when the string makes an angle 4 relative to the vertical. Determine v. Solution: Draw a diagram! Use i to label the initial position of the block and use f to label the final position of the block. Because the tension in the string is perpendicular to the motion at each moment, tension does no work on the block. (The tension is directed radially, towards the centre of the circle on which the block moves, whereas the velocity of the block is always directed tangent to the circle.) Thus, mechanical energy is conserved.

4 By the principle of conservation of mechanical energy, U i + K i = U f + K f K i = U f U i + K f mv i = mgy f mgy i + mv f mv i = mg [y f y i ] + mv f mv i = mg y + mv f vi = g y + vf vi = g y + (0) v = g y In your diagram, sketch a horizontal line through the final position of the block. Use trigonometry in the resulting right triangle to show that y =. ( cos 4 ) Now insert this expression for y into the conservation of mechanical energy equation to obtain an expression for v: v = g(.) ( cos 4 ) v = (9.8)(.) ( cos 4 ) v =.4 m/s The initial speed of the block is.4 m/s. 8. [4 points] Block moves East with speed 8.3 m/s and Block is initially at rest. After the two blocks collide elastically, Block moves 30 North of East. The mass of Block is twice the mass of Block. Determine the speed of Block and the velocity of Block just after the collision. Solution: Draw a diagram! I choose the positive directions to be North and East, which is the usual math-class convention that we have been adopting almost always in lectures. Let v represent the velocity of the first block after the collision and let v represent the velocity of the second block after the collision. Let θ represent the angle of v relative to the East direction, measured South of East. In simplifying the following equations, we make use of the fact that m = m. The x-component of momentum is conserved, so m (8.3) = m v cos 30 + m v cos θ ( ) 3 m (8.3) = m v + m v cos θ ( ) = v + v cos θ (8.3) = 3v + 4v cos θ ()

5 The y-component of momentum is conserved, so 0 = m v sin 30 m v sin θ ( ) 0 = m v m v sin θ 0 = v 4v sin θ v = 4v sin θ () Kinetic energy is conserved, because the collision is elastic, so m (8.3) = m v + m v m (8.3) = m v + m v (8.3) = v + v (8.3) v = v (3) Now we are faced with three equations that we must solve to obtain values for the three unknown quantities, v, v, and θ. Following the strategy we used in lecture, let s first eliminate θ from the first two equations. The best way to do this is to solve equation () for cos θ, solve equation () for sin θ, and then make use of the basic trigonometric identity, cos θ + sin θ = : cos θ = (8.3) 3v 4v sin θ = v 4v Squaring and adding together each side of the previous equation, we obtain ( cos θ + sin (8.3) ) ( ) 3v v θ = + 4v 4v ( ) (8.3) 3v = + v v v ( ) (8.3) 3v + v = v ( v = (8.3) ) 3v + v ( v = 4(8.3) 4(8.3) ) 3v + 3v + v v = 4(8.3) 4(8.3) 3v + 4v 8(v ) = 4(8.3) 4(8.3) 3v + 4v Now substitute for v in the previous equation using the expression from equation (3) to obtain one quadratic equation for v, which can then be placed in standard form

6 and solved using the quadratic formula: 8 [ (8.3) v ] = 4(8.3) 4(8.3) 3v + 4v 8(8.3) 8v = 4(8.3) 4(8.3) 3v + 4v 8(8.3) = 4(8.3) 4(8.3) 3v + v 0 = 4(8.3) 4(8.3) 3v + v 0 = v 4(8.3) 3v 4(8.3) 0 = 3v 8.3 3v (8.3) v = ± (8.3) (3) + 4(3)(8.3) (3) v = ± 3(8.3) + (8.3) v = ± [(8.3) ][3 + ] v = ± (8.3) 5 v = ± v = 8.3[ 3 ± 5] v = 8.3[ 3 + 5] v = 7.75 m/s Note that the negative solution is rejected because we are told the final direction of the first block. Now substitute the value for v into equation (3) to obtain v = (8.3) v v = (8.3) (7.75) v = (8.3) (7.75) (8.3) (7.75) v = v =.09 m/s Finally, to determine the angle θ, substitute the values of v and v into equation ()

7 to obtain v = 4v sin θ sin θ = v 4v sin θ = (.09) sin θ = 0.95 θ = 7.8 Thus, the speed of the first block just after the collision is 7.75 m/s and the velocity of the second block just after the collision is.09 m/s E7.8 S.