A 2200 kg truck traveling north at44 km/h turns east and accelerates to54 km/h.

Question

A 2200 kg truck traveling north at44 km/h turns east and accelerates to54 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° (measuredclockwise 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° (measuredclockwise from east)

Explanation / Answer

    Mass, M = 2200 kg     Initial Velocity, U = 44 km/h                               = 44 * 5 / 18 m/s       = 12.22 m/s     Final velocity, V = 54 km/h                              = 54 * 5 / 18 m/s                              = 15 m/s (a)     Change in KE = ( 1/2 ) M ( V2 -U2)                           = ( 1/2 ) * 2200 * ( 225 - 149.33 )                           = 83238.8 J                           = 83.238 kJ (b)     Change in velocity , V = (V2 + U2 ) ( Here U and V areperpendicular )                                         = 19.35 m/s     Change in the linear momentum, p = MV = 2200 * 19.35 = 42564.7 kg m/s (c)     Direction of change of linear momentum, =arc tan ( U / V )                                                                     = arc tan ( 12.22 / 15 )                                                                     = 39.17 o ( clockwise from East )

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