Question

A 1,140-kg car and a 12,600-kg truck approach each other from opposite directions. The truck is traveling at [01]____________________ km/hr. a. At what speed (m/s) would the car have the same momentum as the truck? b. The car and truck in (a) have a head-on collision and then stick together. What is their final common velocity (m/s)? (Assume the car is going in the positive direction.) c. What is the car’s change in momentum (kg m/s) in (b)? d. At what speed (m/s) would the same car have the same kinetic energy as the truck? e. The car and truck in (d) have a head-on collision and then stick together. What is their final common velocity (m/s)? (Assume the car is going in the positive direction.) f. What is the car’s change in momentum (kg m/s) in (e)?

Answer 1

Answer:

a) $v = 307.02\,\frac{m}{s}$, b) $v=0\,\frac{m}{s}$, c) $\Delta p = - 350002.8\,\frac{kg\cdot m}{s}$, d) $v\approx 92.349\,\frac{m}{s}$, e) $v = - 17.811\,\frac{m}{s}$, f) $\Delta p = - 125582.4\,\frac{kg\cdot m}{s}$

Explanation:

Let assume that truck goes at 100\,km/h (27.778\,m/s).

a) Momentum of the truck is:

$p = (12600\,kg)\cdot (27.778\,\frac{m}{s} )$

$p = 350002.8\,\frac{kg\cdot m}{s}$

The speed needed by the car is:

$v=\frac{350002.8\,\frac{kg\cdot m}{s} }{1140\,kg}$

$v = 307.02\,\frac{m}{s}$

b) The common velocity of the system is given by the application of the Principle of Moment Conservation. Given that both vehicles have momentums with same magnitude but opposite to each other, the total moment of the system is equal to zero and final velocity must be zero.

$v=0\,\frac{m}{s}$

c) The change in momentum is:

$\Delta p = (1140\,kg )\cdot (0\,\frac{m}{s} - 307.02\,\frac{m}{s} )$

$\Delta p = - 350002.8\,\frac{kg\cdot m}{s}$

d) The kinetic energy of the truck is:

$K = \frac{1}{2}\cdot (12600\,kg)\cdot (27.778\,\frac{m}{s} )^{2}$

$K = 4861188.889\,J$

The speed needed by the car is:

$v = \sqrt{\frac{2\cdot (4861188.889\,J)}{1140\,kg} }$

$v\approx 92.349\,\frac{m}{s}$

e) The common velocity of the system is given by the application of the Principle of Moment Conservation.

$(1140\,kg)\cdot (92.349\,\frac{m}{s} )-(12600\,kg)\cdot (27.778\,\frac{m}{s} ) = (13740\,kg)\cdot v$

The final velocity of the system is:

$v = - 17.811\,\frac{m}{s}$

f) The change in momentum is:

$\Delta p = (1140\,kg )\cdot (-17.811\,\frac{m}{s}- 92.349\,\frac{m}{s} )$

$\Delta p = - 125582.4\,\frac{kg\cdot m}{s}$

Answer 2

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