Could someone work this out on paper without i hat or j hat? A 1600 kg car trave

Question

Could someone work this out on paper without i hat or j hat?

A 1600 kg car travelling eastward at 32 m/s collides at an intersection with a 700 kg motorcycletravelling northward at 45 m/s. The two vehicles become entangled after the collision.  

a.) What is the magnitude and direction of the vehicle's velocity after the collision?  

b.) What is the direction of the car's velocity after the collision? (Measured from the east direction)

c.) How much kinetic energy is lost by the two vehicles during the collision?

Explanation / Answer

conservation of momentum

P = 0

mv + MV - (m + M)V' = 0

(1600)(32) {0, 1} + (700)(45){ 1, 0} - (1600 + 700)V' = 0

2300 V'=51200+31500
V' = 82700 /2300
V' = 35.96 m/s
V' = (13.70, 22.26)m/s
magnitude,

|V'|² = 13.7² + 22.26²

|V'| = 26.138 m/s

direction,

tan = 13.7/22.26

tan = 0.6154

= 31.61°

initial kinetic energy,

Eki = ½ m v² + ½ M V²

Eki = ½ (1600)(32)² + ½ (700)(45)²

Eki = 1527.95 KJoule

final kinetic energy,

Ekf = ½ (m + M) V'²

Ekf = ½ (1600 + 700)(26.14)²

Ekf = 785.795 KJoule

lost of kinetic energy,

= Eki - Ekf

= 1527.95 - 785.79
= 742.16 KJoule

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