2211 Physics Spring 2008
Quiz #4
Solutions
Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) Two blocks of mass m1 and m2 are tied together and pulled so that it travels up a frictionless incline that makes an angle with the horizontal, as illustrated. The force pulling them is a tension of magnitude T1 , and their resulting acceleration is not zero. What is the tension magnitude T2 in rope 2, which ties the blocks together, in terms of any or all of m1 , m2 , , T1 , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
There is more than one approach to solving this Newton's Second Law problem. The easiest is probably to look first at the two blocks m1 and m2 along with rope 2 as a single object being pulled up the incline by rope 1. A Free-Body Diagram has been sketched for this object. A coordinate system has been chosen so the acceleration on one axis is zero, and the weight force has been resolved into components. Applying Newton's Second Law to the x direction: Fx = T1 - wtotx = mtot ax ax = T1 - (m1 + m2 ) g sin = (m1 + m2 ) ax
T1 - (m1 + m2 ) g sin m1 + m2
Looking next at just the block with mass m2 , which is pulled up the ramp by rope 2, a Free-Body Diagram has been sketched for this object. The same coordinate system is used, so the acceleration of this block will be represented by the same symbol as the acceleration of the two blocks together. The weight force has been resolved into components. Again, applying Newton's Second Law to the x direction: Fx = T2 - w2x = mtot ax T2 - m2 g sin = m2 ax
Substituting the expression for the acceleration and solving for T2 : T2 - m2 g sin = m2 T1 - (m1 + m2 ) g sin m1 + m2 T2 = m2 T1 - (m1 + m2 ) g sin m1 + m2 + m2 g sin
That is sufficient to answer the question. A bit more algebra, though, yields an interesting result: T2 = m2 = m2 m1 + m2 T1 - (m1 + m2 ) g sin + m2 g sin m1 + m2 m1 + m2 T1 - (m1 + m2 ) g sin + (m1 + m2 ) g sin m2 = m1 + m2 m1 + m2
T1
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II. (16 points) A physics book of mass M is connected by a string to a coffee cup of mass m. The book is given a push up a slope that makes an angle with the horizontal, and released at initial speed vi , as illustrated. The coefficient of static friction between the book and the slope is s , while the coefficient of kinetic friction is k . How much time is required for the book to come to a stop after it is released, in terms of any or all of M , m, , s , k , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
Announced during quiz: your answer may depend on vi . Newton's Second Law will be used for each object to find the acceleration. This acceleration can then be used in a constant-acceleration kinematics expression to find the time for the book to come to rest. The acceleration magnitude of the two objects is the same, and the tension in the string between the two objects is the same on each object. Looking first at the cup, a Free-Body Diagram has been sketched. A coordinate system has been chosen so that the acceleration is positive (the cup slows as it rises, so the acceleration is downward). Applying Newton's Second Law to the cup and solving for the tension: Fx = wc - T = max T = mg - max
Turning now to the book, a Free-Body Diagram has been sketched. A coordinate system has been chosen so that the acceleration is positive (the book slows as it ascends the slope, so the acceleration is down the slope) along one axis and zero along the other, making the acceleration symbolized by ay for the book the same as that symbolized by ax for the cup. The weight of the book has been resolved into components. The normal force on the book can be found by applying Newton's Second Law in the z direction: Fz = n - wbz = M az = 0 n = M g cos
Next, Newton's Second Law will be applied in the y direction: Fy = T + fk + wby = M ay T + k n + M g sin = M ay
Substituting the expressions for the tension and the normal force: (mg - max ) + k (M g cos ) + M g sin = M ay Note that ax = ay . Let them both be "a", and solve for it: mg + k M g cos + M g sin = M a + ma a= m + M (sin + k cos ) g M +m
Since the distance the book slides is neither known nor asked, the "no displacement" constant-acceleration kinematics equation, v = v0 + a (t) can be used to find the time to slide to a stop. When it stops, the velocity v is zero. Since the positive direction has already been chosen to be down the incline, the initial velocity is negative, so v0 = -vi . 0 = -vi + a (t) t = vi a t = vi (M + m) (m + M (sin + k cos )) g
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1. (6 points) In the problem above, while the book is sliding up the ramp, how does the magnitude of the tension in the string, Ts , compare to the weight of the cup, Wc ? . . . . . . . . . . . . . . . . . . . . . . . Since the cup comes to a stop as it rises, the acceleration is positive downward. Therefore, the net force must be downward, and Ts < Wc
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III. (16 points) Far out in space where gravity is negligible, a spaceship of mass m traveling at a positive velocity v0 fires a combination of its engines and retrorockets, beginning at time t = 0. The resulting thrust is illustrated in the graph a as function of time. Note that the maximum thrust in the negative direction has magnitude Fneg , and the maximum thrust in the positive direction (reached at time t = t2 ) has magnitude Fpos . What impulse is imparted to the spaceship between time t = 0 and time t = t4 , in terms of any or all of m, v0 , t1 , t2 , t3 , t4 , Fpos , Fneg , and physical or mathematical constants? . . . . . . . . . . . . . . . . . . . . . . . Since J = F dt, the impulse is the area under the graph of force as a function of time. Remember that the area of a triangle is one-half its base times its height.
1 2
- Fneg (t1 - 0) + Fpos (t3 - t1 ) - Fneg (t4 - t3 )
Note: Since the problem statement defines Fneg as a magnitude, but the graph strongly implies it is a signed number, full credit was also issued for:
1 2
Fneg (t1 - 0) + Fpos (t3 - t1 ) + Fneg (t4 - t3 )
2. (6 points) Assuming the graph in the problem above accurately depicts relative magnitudes (that is, Fpos > Fneg , (t3 - t2 ) > (t4 - t3 ), etc.), at what time does the spaceship reach its maximum speed? . . . . . . . . . . . . . . . . . . . . . . . Since the initial velocity v0 is positive, positive forces and the resulting positive accelerations will increase the speed, while negative forces will decrease it. There are no positive forces after t = t3 , so the spaceship cannot travel any faster than it does at that time. The maximum speed is reached at t = t3
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3. (8 points) A ball of clay is thrown horizontally toward a block sitting at rest on a horizontal frictionless surface. The clay hits and sticks to the block. What is true about the magnitude of the ball's momentum before it hits the block, pball , and the magnitude of the impulse exerted by the ball on the block, Jball on block ? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . After the clay ball hits and sticks to the block, they will slide together on the frictionless surface in the same direction that the clay ball was thrown. The change in the block's momentum, and thus the impulse on it, is in the same direction as the ball's original momentum. Because the ball is still travelling in its original direction after it hits the block, the magnitude of its momentum change was less than the magnitude of its original momentum. The magnitude of the ball's momentum change is the impulse on it, which must be equal in magnitude (although opposite in direction) to the impulse on the block. The impulse on the block, then, must also have a magnitude that is less than the ball's original momentum. pball > Jball on block , and they are in the same direction.
4. (8 points) A rubber ball and a clay ball with equal masses are thrown directly at a wall with equal speeds. The rubber ball bounces, but the clay ball sticks. Which ball, if either, exerts an impulse of greater magnitude on the wall? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The rubber ball has twice the momentum change, because it changes direction (momentum is a vector). This momentum change is the impulse on the ball. The impulse on the wall is equal in magnitude to the impulse on the ball (although opposite in direction). The rubber ball exerts an impulse of greater magnitude because it bounces.
5. (8 points) A small car is pushing a larger truck with a dead battery, causing it to speed up to the right, as illustrated. The mass of the truck is greater than that of the car. Which of the following statements is true? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The force of the car on the truck and the force of the truck on the car constitute an action/reaction force pair. The car exerts the same magnitude force on the truck as the truck exerts on the car.
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6. (8 points) Boxes A and B are sliding to the left across a frictionless table. The hand H is slowing them down. The mass of A is less than the mass of B. Rank the magnitudes of the horizontal forces on A, B, and H, from greatest to least. (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . The blocks have the same acceleration. If they are considered together as one object, then the force from the hand, FH on A , is the only external force on that object, and must be the cause of the acceleration. If box B is considered individually, then the external force on it, which causes the same acceleration, is the force from box A, FA on B . Since box B by itself must have less mass than boxes A and B together, FA on B < FH on A . Forces that constitute action/reaction pairs must have equal magnitude, so FA on H = FH on A and FB on A = FA on B . Putting these relationships together yields FA
on H
= FH
on A
> FB
on A
= FA
on B
7. (8 points) A book is at rest on a level table top. What force, if any, is part of an action/reaction pair (or "force pair") with the weight of the book? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Every force is part of an action/reaction pair. The forces which constitute a particular action/reaction pair represent the interaction of two objects, and so must be on two different objects. They must be the same phenomenon. Since the weight of the book is a gravitational force from the Earth on the book, the other force in the pair must be The gravitational force of the book on the Earth.
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