A 2000 kg truck traveling north at39 km/h turns east and accelerates to52 km/h.

Question

A 2000 kg truck traveling north at39 km/h turns east and accelerates to52 km/h. (a) What is the change in the truck's kineticenergy?
1 J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
2 kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
3° (measured clockwise from east) (a) What is the change in the truck's kineticenergy?
1 J

(b) What is the magnitude of the change in the linear momentum ofthe truck?
2 kg·m/s

(c) What is the direction of the change in the linear momentum ofthe truck?
3° (measured clockwise from east)

Explanation / Answer

   v = 52 km/h = 52*5/18 m/s = ... m/s    u = 39 km/h = 39*5/18 m/s = ... m/s    From now on use u and v only in m/s form.   (a) Change in kinetic energy =1/2*m*[v2 - u2] = 1000 * [v2 -u2] = ...... J (b) suppose we denote unit vectors along north and eastby j and i respectively,        change in linea momentum= m * [ v i - u j] = 2000 * [ v i + u j ]        Now, i and j areperpendicular to each other.        So, to take the magnitudetake v = [v*v - u*u ] = ..... m/s        So, magnitude of changein linear momentum = mv = ..... kgm/s (c) direction of change in linear momentum =tan-1(mu/mv)    = tan-1(u/v) clockwise from east    = 36.87 degrees CW from east

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