Solutions for Problem Set 9

Last revised 1999/11/02

Chapter 9

11.

(MathCad solution)

16.
Let the position of the CM of the car+cannon be y before the cannonballs are shot and y-x afterwards, and put the origin of the coordinate system at the left end of the car. Then the position of the CM before firing is

y_CM = m*0 + M y

and after all the firing is over, noting that the car has moved backwards an amount x, the CM is

y_CM = m (L - x) + M (y-x)

Since there are no external forces on the system, the center of mass cannot have moved, and we have

m*0 + M y = m (L - x) + M (y-x), giving

x = m L / (M + m)

If m=0 the car can't move, and indeed in that case x = 0. If M=0, x = L, which is the largest amount the car could possibly move.

40.
The easiest (only) thing we can do here is to use momentum conservation. The total momentum before the decay is zero. Taking  y along  y_nu and x along  -P_e, the momentum afterwards is

P = (-P_e + P_Z,x ) 1_x + (P_nu - P_Z,y) 1_y = 0

where the "=0" comes from momentum conservation. Clearly the x-component of P_Z is P_e and its y-component is  -P_nu . From the properties of right triangles,

|P_Z| = sqrt (P_e ² + P_nu ² )    and    theta_P =  atan(P_nu / P_e )

47.
(MathCad solution)

55.

From the rocket equation,

m(t) a = v_ex (dm/dt), where v_ex < 0

Integrating with respect to t gives

integral(a dt) = v_ex integral(1/m dm/dt dt) = integral(1/m dm)

v_f - v₀ = v_ex ln (m/m₀₎

and since the final velocity of the rocket is supposed to be -v_ex in order that the exhaust gasses be at rest, while the initial velocity of the rocket is zero,

- v_ex - 0 = v_ex ln (m/m₀₎

Dividing through by v_ex and taking the exponential of both sides gives

e^-1 = m/m₀

m = m₀/e

Links to image files used in these solutions: 11, 11a, 11b, 11c, 11d, 11e, 11f, 11g; 16; 40;