q7

Quiz 7

11/23/2008 03:25:59 PM. Note correction to #34 ! : 2/5 of the total energy goes to the piece with mass (3/2)m.
3/5 of the total energy goes to the smaller piece.

Ch. 7:  elastic collisions: 22, 23, 24, 27, 28  inelastic collisions: 4, 6, 8,  31, 32, 33, 34, 35   
TURN IN:
4, 6, 24, 31, 35

22.

You have two equations and two unknowns:

m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 .

and

v_i1 - v_i2 = +v_f2 - v_f1 .

The unknowns are v_f2 and v_f1 .

Substitute the values of the masses and given velocities into the equations to find the final velocities.

Let rightward (EAST) be the positive direction. The initial velocity of mass 1 is 3.30 m/s; mass 1 = 0.440 kg. The initial velocity of mass 2 is zero; mass 2 = 0.220 kg. Plug these values into the first equation and the second equation:
v_i1 - v_i2 = +v_f2 - v_f1, which becomes,

3.30 m/s - 0 = +v_f2 - v_f1 .

Solve the two equations for the two unknowns

23. Same problem as the previous, except that the first mass will be going in the negative direction after the collision.

You have two equations and two unknowns:

m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 .

and

v_i1 - v_i2 = +v_f2 - v_f1 .

The unknowns are v_f2 and v_f1 .

Substitute the values of the masses and given velocities into the equations to find the final velocities.

Let rightward (EAST) be the positive direction. The initial velocity of mass 1 is 3.00 m/s; mass 1 = 0.450 kg. The initial velocity of mass 2 is zero; mass 2 = 0.900 kg. Plug these values into the first equation and the second equation:
v_i1 - v_i2 = +v_f2 - v_f1, which becomes,

3.00 m/s - 0 = +v_f2 - v_f1 .

Solve the two equations for the two unknowns. Hint: The final velocity of the first mass will be negative.

24.

Same problem as the previous two, except that the masses are initially going in opposite directions. Also the masses are identical, so they cancel out in the first equation.

You have two equations and two unknowns:

m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 .

and

v_i1 - v_i2 = +v_f2 - v_f1 .

The unknowns are v_f2 and v_f1 .

Substitute the values of the given velocities into the equations to find the final velocities.

Let rightward (EAST) be the positive direction. The initial velocity of mass 1 is 2.00 m/s. The initial velocity of mass 2 is -3.00 m/s. Plug these values into the first equation and the second equation:
v_i1 - v_i2 = +v_f2 - v_f1, which becomes,

2.00 m/s - (-3.00 m/s) = 5.00 m/s = +v_f2 - v_f1 .

Solve the two equations for the two unknowns. Hint: The final velocity of the first mass will be negative. The final velocity of the second mass will be positive. Indeed, the two balls will exchange velocities.

27.

Same problem as the previous ones, except that the masses are initially going in the same direction.

You have two equations and two unknowns:

m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 .

and

v_i1 - v_i2 = +v_f2 - v_f1 .

The unknowns are v_f2 and v_f1 .

Substitute the values of the masses and given velocities into the equations to find the final velocities.

Let rightward (EAST) be the positive direction. The initial velocity of mass 1 is 4.50 m/s. The initial velocity of mass 2 is 370 m/s. Plug these values into the first equation and the second equation:
v_i1 - v_i2 = +v_f2 - v_f1, which becomes,

4.50 m/s - (3.70 m/s) = 0.80 m/s = +v_f2 - v_f1 .

Solve the two equations for the two unknowns.

28.

Same problem as the previous ones, except that you have different unknowns: the initial velocity of the first mass and the value of the second mass. You want the second mass value.

You have two equations and two unknowns:

m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 .

and

v_i1 - v_i2 = +v_f2 - v_f1 .

The unknowns are v_i1 and m₂.

Substitute the values and conditions given to find the unknowns.

Let rightward (EAST) be the positive direction. The initial velocity of mass 1 is v_i1; mass 1 is 0.280 kg. The initial velocity of mass 2 is ZERO; mass 2 has value m₂ . Plug these values into the first equation and the second equation:
v_i1 - v_i2 = +v_f2 - v_f1, which becomes,

v_i1 - v_i2 = +v_i1/2 - v_f1 , or

v_i1 = +v_i1/2 - v_f1 .

Thus the two solvable equations are:

(A) m₁v_i1 = m₁v_f1 + m₂v_i1/2 .

and

v_i1 = +v_i1/2 - v_f1 or

(B) v_f1= - v_i1/2.

Solve the two equations for m₂ . Just take the expression for v_f1 in equation (B) and substitute into equation (A). v_i1will cancel out. Find the value of the second mass given that m₁ = 0.280 kg.
(b) In this case you must simply compare the kinetic energy of the croquet ball before and after the collision by taking the difference of the two. Then divide this difference by the kinetic energy of the croquet ball before the collision. For that ball: the kinetic energy before the collision is (1/2)m₁v_i1². Yo, after the collision, the kinetic energy is (1/2)m₁v_i1²/4, from equation (B) above. Take the difference of these two expressions and divide by the kinetic energy of the croquet ball before the collision. Your answer will be a simple decimal fraction.

4. m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 only; no second equation needed. The initial momentum is zero.
0 = m₁v_f1 + m₂v_f2 .
See example 7.4.

8. See example 7.3. But in this case you are solving for m_B.

6. See the first equation stated in problem 4, at the top. m₁v_i1 + m₂v_i2 = m₁v_f1 + m₂v_f2 . Let 1 indicate the half back and 2 the cornerback. Plug the given initial values of the velocity into the equation. The players have the same final velocity V_f. Plug in this symbol into the equation. Solve for V_f.

31. and 32. See example 7.10. For #32, you have to decompose the displacement along the arc into horizontal and vertical components. But the vertical component is h. So now you must find x, the horizontal component, using the length L of the pendulum rod. It can be shown, and discovered with a little thought, that the following equation is true: L² = x² + (L - h)². Now L is given, and h you can find from example 7.10. Solve for x.

33. See example 7.10. The kinetic energy just after the collision is given by the expression on the left hand side of equation (ii). On the next line down, the speed v' of the system (bullet + block) just after the collision is given in terms of the height h. The kinetic energy of the bullet before the collision is given by (1/2)mv². The speed v is given on the last line in terms of h. Subtract the two expressions to evaluate (bullet + block kinetic energy just after) - (bullet kinetic energy before). Divide this difference by (bullet kinetic energy before). The height h will cancel out.

34.hint on this later. This is an easy problem, it's like dividing up a pie into two two pieces once we know the speed of each particle of given mass. Let m be the mass of the first piece ; (3/2)m is the mass of the other. Conservation of momentum gives:
mV = -(3/2)mV'.
where the two velocities are noted.
Thus V' = -(2/3)V .
Now the kinetic energy is,
(1/2)mV² + (1/2)(3/2)mV'² .Substituting the expression for V' in terms of V we get;

(1/2)mV² + (1/2)(3/2)(4/9)mV² = 7500 J = total energy

or

(1/2)mV² + (1/2)(2/3)mV² = 7500 J. Thus,
2/5 of the total energy goes to the piece with mass (3/2)m.
3/5 of the total energy goes to the smaller piece.
The 2/5 and 3/5 fractions arise from the fact that we are using (1/2)mV² as the basic unit. The second term on the right hand side of the previous equation has a factor of 2/3 in front, suggesting that the first term has a factor of 3/3 in front.
So we have 5 parts of 1/3 each.

35. Let m be the sports car mass and M be SUV mass.
We have mV + M�0 = mV_f + MV_f , for conservation of momentum . V is the initial velocity of the sports car and V_f is the final common velocity of the locked cars just after the collision. That is :

mV = (m + M)V_f. Find the final common velocity just after the collision.

Now, after the collision use conservation of energy:
KE_f + PE_f = KE_ff + PE_ff+ HEAT .
(1/2)(M + m)V_f² + PE_f = KE_ff + PE_ff+ HEAT .

"f" refers to just after collision and "ff" to when the locked cars comes to rest after sliding down the street and generating HEAT.

At "ff", the kinetic ( at rest) and potential (at y = 0) energies are zero. Note the y = 0 convention makes PE_f = 0. Thus:

(1/2)(M + m)V_f² + 0 = 0 + 0 + HEAT .
(1/2)(M + m)V_f² = HEAT .

HEAT = f_kD = �(m + M)gD.

D is given. Evaluate the HEAT and work backwards to find the initial speed V .

more hint on this to come !

More hints soon