50:750:131, Elements of Physics Name ________________

Test 2. November 10, 2004

You must show the working for all solutions. No credit will be given for an answer without the work. Take g to be 9.80 m/s2

1. An 80 kg block is pulled at a constant speed of 4.0 m/s across a horizontal floor by an applied force of 120 N directed 40 above the horizontal. What is the rate at which the force does work on the block?

The horizontal component of the force is Fx = (120 N)cos(40) = 91.925 N

The rate at which it does work, or the power, is P = Fx v = (91.925 N)(4.0 m/s) = 367.70W

(More compactly, P = F.v)

2. A block of mass m = 2.5 kg is dropped from height h = 40 cm onto a spring of spring constant k = 2000 N/m. Find the maximum distance the spring is compressed. (Use figure 8-36)

The block initially has zero kinetic energy. It falls a distance h onto the spring, but, when it reaches the spring, it is moving at some speed, so it continues downwards, compressing the spring a distance x. It eventually slows down and stops. Of course, it then bounces back up again, but at the instant of time when it stops, it has no kinetic energy, it has lost potential energy mg(h + x) and this has been converted into elastic potential energy of the spring, ½ kx2.

½ kx2 = mg(h + x)

½ kx2 -mgx - mgh = 0

Putting in numbers

1000 x2 - (2.5*9.8)x - (2.5*9.8*0.40) = 0

This is a quadratic equation. One solution is positive (that is the one we need) and the other is negative (what does that mean?). The positive solution is

x = 0.112 m = 11.2 cm

3. A 1000 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of 5.0 m/s2. At the same instant a 2000 kg truck, traveling at a constant speed of 10.0 m/s, overtakes and passes the automobile. (a) How far is the center of mass of the automobile - truck system from the traffic light at t = 3.0 s? (b) What is the speed of the center of mass then?

At t = 3.0 s, the position of the automobile is x1 = ½ at2 = 22.5 m

and its velocity is v1 = at = 15.0 m/s

The truck moves at constant velocity v2. At t = 3.0 s is position is x2 = v2t = 30 m

and its speed is 10.0 m/s

(a) The position of the center of mass is

xCofM = (m1x1 + m2x2)/(m1 + m2) = (1000*22.5 + 2000*30)/(1000 + 2000) = 27.5 m

(b) The velocity of the center of mass is

vCofM = (m1v1 + m2v2)/(m1 + m2) = (1000*15.0 + 2000*10.0)/(1000 + 2000) = 11.67 m/s

4. What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius 3600 km at a speed of 25,000 km/h?

(a) w = v/r = 25,000/3600 = 6.944 rad/h = 1.929 × 10-3 rad/s

(b) aradial = -v2/r = (25,000)2/3600 = 1.7361 × 105 km/h2 = 13.396 m/s2

(c) Since the speed of the spaceship is constant, the tangential acceleration is zero.

5. A man stands on a platform that is rotating (without friction) with an angular speed of 1.4 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 5.0 kg-m2. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kg-m2, what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

(a) Angular momentum is conserved, so

Iiwi = Ifwf

wf = wi(Ii/If) = (1.4 rev/s)(5.0 / 2.0) = 3.5 rev/s

(b) The rotational kinetic energy in each case is ½ Iw2 so the ratio of the new kinetic energy to the old is

(2.0 * 3.52)/(5.0*1.42) = 2.5

(There was no need to convert rev/s to radians/s in this case, but, if you were not sure that you could do it this way, there would be no harm in doing the conversion)

(c) Where did the extra energy come from? The man provided it by muscle power. When he pulled the bricks inwards, he would have felt a force resisting him, and he did work overcoming that force.

6. A uniform cubical crate is 0.80 m on each side and weighs 600 N. It rests on a floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude 400 N be applied for the crate to tip?

The small obstruction is at point A. The force F causes the crate to rotate clockwise about point A. Once the crate begins to rotate, the only point on the crate that is in contact with the floor is the corner at A, so the normal force must act there. If we choose A as the axis, and calculate the torques about that point, the only forces that contribute are the weight W and the applied force F.

W * (0.40 m) - F * x

If x is just large enough for the crate to tip, this must be infinitesimally different from zero.

W * (0.40 m) - F * x = 0 if x = W*(0.40m)/F = 600N * 0.40m / 400N = 0.60 m.

7. A body of mass 4.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed pf the 4.0 kg body was 5.0 m/s?

(a) In an elastic collision, v1f = vi(m1 - m2)/(m1 + m2)

If the speed of particle 1 is reduced by a factor of 4

1/4 = (m1 - m2)/(m1 + m2)

4(m1 - m2) = (m1 + m2)

3m1 = 5m2

m2 = 0.6 m1 = 2.4 kg

(b) Before the collision, only mass 1 is moving, at 5.0 m/s, so the speed of the two-body center of mass is (see problem 3 for formula)

vCofM = (4.0 * 5.0 + 2.4 * 0)/(4.0 + 2.4) = 3.125 m/s

With the values used in the quiz:

(a) In an elastic collision, v1f = vi(m1 - m2)/(m1 + m2)

If the speed of particle 1 is reduced by a factor of 5

1/5 = (m1 - m2)/(m1 + m2)

5(m1 - m2) = (m1 + m2)

4m1 = 6m2

m2 = 0.667 m1 = 4.0 kg

(b) Before the collision, only mass 1 is moving, at 3.0 m/s, so the speed of the two-body center of mass is (see problem 3 for formula)

vCofM = (6.0 * 3.0 + 4.0 * 0)/(6.0 + 4.0) = 1.8 m/s

8. A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by 1.0 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

(a) Extract from the information the idea that the father's kinetic energy is doubled when his speed in increased by 1.0 m/s.

½ mf(vf + 1.0)2 = 2 × ½ mfvf2

(vf + 1.0)2 = 2vf2

Take the square root of both sides

vf + 1.0 = 1.4142 vf

0.4142 vf = 1.0

vf = 2.4142

(b) The second piece of information is that the son has half the mass of the father but has, initially, twice the kinetic energy of the father. Therefore

½ (mf/2)vs2 = 2 × ½ mfvf2

1/4 vs2 = vf2

vs2 = 4vf2

vs = 2vf = 4.8284 m/s