imported>Patrick: Created page with ''''(a)''' The momentum of the truck, $p$ , is: $p = m v = (10^4)(30) = 3 \times 10^5 \tfrac{kg m}{s}$

The momentum of the car, $p'$ , is …'

2011-08-30T19:03:44Z

Created page with ''''(a)''' The momentum of the truck, <math>p</math>, is: <math>p = m v = (10^4)(30) = 3 \times 10^5 \tfrac{kg m}{s}</math> The momentum of the car, <math>p'</math>, is …'

New page

'''(a)''' The momentum of the truck, <math>p</math>, is:

<math>p = m v = (10^4)(30) = 3 \times 10^5 \tfrac{kg m}{s}</math>
 
The momentum of the car, <math>p'</math>, is equal to the momentum of the truck, <math>p</math>:

<math>p' = p = 3 \times 10^5 = m' v' = (1.2 \times 10^3) v'</math>
 
Solve for <math>v'</math>:

<math>v' = \frac{3 \times 10^5}{1.2 \times 10^3} = 2.5 \times 10^2 m/s</math>
 
'''(b)''' The kinetic energy of the truck, <math>K</math>, is:

<math>K = \frac{1}{2} m v^2 = \frac{1}{2}(10^4)(30)^2 = 4.5 \times 10^6 J</math>
 
The kinetic energy of the car, <math>K'</math>, is equal to the kinetic energy of the truck, <math>K</math>:

<math>K' = K = 4.5 \times 10^6 = \frac{1}{2} m' {v'}^2 = \frac{1}{2}(1.2 \times 10^3){v'}^2</math>
 
Solve for <math>v'</math>:

<math>v' = \sqrt{2 \times 4.5 \times 10^6 \over 1.2 \times 10^3} = 86.6 m/s</math>

Conservation of Momentum EX 4 - Revision history

imported>Patrick: Created page with ''''(a)''' The momentum of the truck, p, is: p = m v = (10^4)(30) = 3 \times 10^5 \tfrac{kg m}{s} The momentum of the car, p', is …'

imported>Patrick: Created page with ''''(a)''' The momentum of the truck, $p$ , is: $p = m v = (10^4)(30) = 3 \times 10^5 \tfrac{kg m}{s}$

The momentum of the car, $p'$ , is …'