Chapter 7 Quick Check Problems

Quick Check 7.1 A ball rolls around a circular track with an angular velocity of 4 rad/s. What is the period of the motion? s 1 s 2 s Answer: A Reason: T = 2π/ω = 2π / 4π = 1/2

Quick Check 7.2 Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s angular speed is ______ that of Rasheed. Half The same as Twice Four times We can’t say without knowing their radii. Answer: B Reason: This is a rigid (solid) object. All points, regardless of location on the object, will have the same angular speed.

Quick Check 7.3 Rasheed and Sofia are riding a merry-go-round that is spinning steadily. Sofia is twice as far from the axis as is Rasheed. Sofia’s speed is ______ that of Rasheed. Half The same as Twice Four times We can’t say without knowing their radii. Answer: C Reason: This one asks about the SPEED, not angular speed. In this case, we may invoke the relation v = ωr. Point s has a greater radius and so the speed must be greater. In this case, if you double r then you double v.

Quick Check 7.5 The fan blade is slowing down. What are the signs of ω and ? ω is positive and  is positive. ω is positive and  is negative. ω is negative and  is positive. ω is negative and  is negative. ω is positive and  is zero. Answer: C Reason: The blade is rotating in the clockwise direction and so its angular speed is negative. Since it is slowing down, the acceleration points in the opposite direction (and thus is positive).

Quick Check 7.6 The fan blade is speeding up. What are the signs of  and ?  is positive and  is positive.  is positive and  is negative.  is negative and  is positive.  is negative and  is negative. Answer: D Reason: The blade is moving in the clockwise direction and so the angular speed is negative. It is speeding up, so the acceleration points in the same direction (and thus is also negative). 6

Quick Check 7.8 Starting from rest, a wheel with constant angular acceleration spins up to 25 rpm in a time t. What will its angular velocity be after time 2t? 25 rpm 50 rpm 75 rpm 100 rpm 200 rpm Answer: B Reason: α = Δω / Δt  Δω = αΔt. Thus, if we double Δt we will also double Δω from 25 to 50 rpm.

Quick Check 7.10 The four forces shown have the same strength. Which force would be most effective in opening the door? Force F1 Force F2 Force F3 Force F4 Either F1 or F3 Answer: A Reason: We are looking for which force would supply the greatest torque. Force 2 would supply no torque as it points along the radial line. We know that τ = rF̝ so the further away the force is from the hinge, the greater the torque it will apply. We then know that force 1 is better than force 4. Force 1 is also greater than force 3 since 4 is entirely perpendicular to the door. This, force 1 must supply the greatest torque and is the most efficient at opening the door. 8

Stop to Think 7.3 A wheel turns freely on an axel on the center. Which of the given forces will provide the largest positive torque on the wheel? A B C D E Answer: A Reason: C and D yield no torque at all since they point along the radial line. A and B cause the object to rotate clockwise (the negative direction). This, the only force that gives any positive torque is that of E. 9

Chapter 9 Quick Check Problems

Stop to Think 9.1 Which force delivers a greater impulse? Force A Force B Both deliver the same impulse Answer: C Reason: J = area under the curve of a force vs. time graph. In these cases, the area is simply length x width. On the left, we have (300 N)*(0.005 s) = 1.5 kgm/s. On the right we have (150 N)*(0.010 s) = 1.5 kgm/s. They are identical so the correct answer is C.

Quick Check 9.2 A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? 0.50 m/s left At rest 0.50 m/s right 1.0 m/s right 2.0 m/s right Answer: D Reason: J = Δp = pf – pi = mvf – mvi = m(vf – vi)  vf = J/m + vi = area/m + vi = (2N)(0.5s)/2kg + 0.5m/s = 0.5 + 0.5 = 1.0 m/s. Positive, so to the right. 12

Quick Check 9.3 A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends? 0.50 m/s left At rest 0.50 m/s right 1.0 m/s right 2.0 m/s right Answer: B Reason: J = Δp = pf – pi = mvf – mvi = m(vf – vi)  vf = J/m + vi = area/m + vi = (-2N)(0.5s)/2kg + 0.5m/s = -0.5 + 0.5 = 0.0 m/s. No speed so it is at rest. 13

Quick Check 9.5 A light plastic cart and a heavy steel cart are both pushed with the same force for 1.0 s, starting from rest. After the force is removed, the momentum of the light plastic cart is ________ that of the heavy steel cart. Greater than Equal to Less than Can’t say. It depends on the size of the force. Answer: B Reason: The same force is applied over the same amount of time. As a result, the impulse is the same (J=FΔt). Furthermore, since J=Δp, the change in momentum must also be the same. 14

Quick Check 9.1 The cart’s change of momentum px is –20 kg m/s Answer: E Reason: Δp = pf – pi = mvf – mvi = 10 kgm/s – (-20 kgm/s) = 30 kgm/s. 15

Quick Check 9.7 Throw the Superball. Throw the ball of clay. You awake in the night to find that your living room is on fire. Your one chance to save yourself is to throw something that will hit the back of your bedroom door and close it, giving you a few seconds to escape out the window. You happen to have both a sticky ball of clay and a super-bouncy Superball next to your bed, both the same size and same mass. You’ve only time to throw one. Which will it be? Your life depends on making the right choice! Throw the Superball. Throw the ball of clay. It doesn’t matter. Throw either. Answer: A Reason: The superball will bounce off of the wall and the clay will stick. Since the ball bounces back, it will have a greater change in momentum than the ball of clay that simply sticks to the wall. A larger Δp means a larger impulse (J=Δp) and, therefore, a larger applied force. 16

Quick Check 9.8 A mosquito and a truck have a head-on collision. Splat! Which has a larger change of momentum? The mosquito The truck They have the same change of momentum. Can’t say without knowing their initial velocities. Answer: C Reason: Momentum is conserved! There wi have just the forces between the two objects (no external forces to worry about). Δpbug + Δptruck = 0. 17

Quick Check 9.10 The two boxes are on a frictionless surface. They had been sitting together at rest, but an explosion between them has just pushed them apart. How fast is the 2-kg box going? 1 m/s 2 m/s 4 m/s 8 m/s Not enough information. Answer: B Reason: pf = pi. But the initial momentum is zero since it is at rest, so: pf = m1v1f + m2v2f = 0  v2f = -(m1/m2)*(v1f) = -1/2 * -4 = 2 m/s. 18

Quick Check 9.11 The 1-kg box is sliding along a frictionless surface. It collides with and sticks to the 2-kg box. Afterward, the speed of the two boxes is 0 m/s 1 m/s 2 m/s 3 m/s There’s not enough information to tell. Answer: B Reason: pf = pi  (m1 + m2)vf = m1v1i + m2v2i. But v2i is zero and after rearranging we have: vf = m1v1i/(m1 + m2) = 1 m/s. 19

Chapter 10 Quick Check Problems (coming soon)