Question

Two cars are approaching a perpendicular intersection without a stop sign. Car 1 has a mass m1 =700kg and is heading north at v2 =15m/s. Car 2 has mass m2 =600kg and is heading west at an unknown speed v2. The two cars collide at the intersection and stick together as a result of the collision. The police report stated that after the collision, the two cars moved together in a direction 40° west of north and stopped after sliding d =35m from the collision point. 

Part 1: Find the speed of car 2 before the collision (3 pt). 

Part 2: What is the speed of the cars immediately after the collision? (3 pt) v= 

Part 3: What is the coefficient of kinetic friction between the cars and the pavement? (2 pt) Mr = 

Part 4: What is the magnitude of the impulse exerted on car 1 during the collision? (2 pt) J1 =

Answer 1

2 1) 700 X 15 =8.07 69 mp (700 +600) is rls C 40° v Cos yo = 8.0769 3 с Love 10.J4 367 ms v= VW lo-54 367x Sin 40" 6.777 m/s S ₂ (700+600) (6.777) (600) 14.684 mps V 10.54 m/s » Immediately offer collision, 3 (10.54362 a = ( =-1.588 m/8² 2x 35 eg - a Moe = 1.588 9.8 0.162 وا Impulse best 700 (-10.543+15) 3119.43 kg mlo

V= Vm3v3 + m?v? (m+m2) (6002)2+(700x15) 700+600 vsina 600v2 700 x 15 VCOSC mi 2v2 tan40 = 35 V2 35 2 x tan40 = 14.684 x m/s Therefore speed of car 2 before collision 14.684 x m/s Now V(600v2 ) + ( 700x 15) v= 700+600 (600 x 14.684)2 + ( 700 x 15) V= 700+600 v = 10.544 x m/s Speed of the cars immediately after collision 10.544 x m/s Cars stopped after d = 35 m v = v2 + 2ad 0 = (10.544)2 +2 * (- 448) 0 = (10.544)2 + 2 x (-4x) * (9.8) X (35)

Mk (10.544) 2 (2) * (9.8) * (35) 0.162 Impulse exerted on car 1 1,-së, --11:51 - 1,= m { vsinai + vcosaj) - vj} 1, = m; V(usina)2 + (vcosa - vi) 2 J. = (700) « V<(10.544) * (sin40)}2 + {(10.544 x cos40) - (15)}2 Jy = 6781.7kg · m/s

Answer 2

2 1) 700 X 15 =8.07 69 mp (700 +600) is rls C 40° v Cos yo = 8.0769 3 с Love 10.J4 367 ms v= VW lo-54 367x Sin 40" 6.777 m/s S ₂ (700+600) (6.777) (600) 14.684 mps V 10.54 m/s » Immediately offer collision, 3 (10.54362 a = ( =-1.588 m/8² 2x 35 eg - a Moe = 1.588 9.8 0.162 وا Impulse best 700 (-10.543+15) 3119.43 kg mlo

V= Vm3v3 + m?v? (m+m2) (6002)2+(700x15) 700+600 vsina 600v2 700 x 15 VCOSC mi 2v2 tan40 = 35 V2 35 2 x tan40 = 14.684 x m/s Therefore speed of car 2 before collision 14.684 x m/s Now V(600v2 ) + ( 700x 15) v= 700+600 (600 x 14.684)2 + ( 700 x 15) V= 700+600 v = 10.544 x m/s Speed of the cars immediately after collision 10.544 x m/s Cars stopped after d = 35 m v = v2 + 2ad 0 = (10.544)2 +2 * (- 448) 0 = (10.544)2 + 2 x (-4x) * (9.8) X (35)

Mk (10.544) 2 (2) * (9.8) * (35) 0.162 Impulse exerted on car 1 1,-së, --11:51 - 1,= m { vsinai + vcosaj) - vj} 1, = m; V(usina)2 + (vcosa - vi) 2 J. = (700) « V<(10.544) * (sin40)}2 + {(10.544 x cos40) - (15)}2 Jy = 6781.7kg · m/s

Answer 3

2 1) 700 X 15 =8.07 69 mp (700 +600) is rls C 40° v Cos yo = 8.0769 3 с Love 10.J4 367 ms v= VW lo-54 367x Sin 40" 6.777 m/s S ₂ (700+600) (6.777) (600) 14.684 mps V 10.54 m/s » Immediately offer collision, 3 (10.54362 a = ( =-1.588 m/8² 2x 35 eg - a Moe = 1.588 9.8 0.162 وا Impulse best 700 (-10.543+15) 3119.43 kg mlo

Answer 4

(a)

conserving momentum in x-direction

600 * -v2 = ( 700 + 600) * - v * sin 40

conserving momentum in y-direction

700 * 15 = ( 700 + 600) * v * cos 40

solve the second equation for v, we get

v = 10.54 m/s

so,

v2 = 14.68 m/s

_____________________

(b)

speed after collision = 10.54 m/s

________________________

(c)

v² - u² = 2ad

v² - u² = -2ugd

10.54² = 2 * u * 9.8 * 35

solve for u, i got

u = 0.162

__________________-

(d)

impuse = change in momentum

impulse = 700 ( 10.54 - 15)

impulse = 3122 Kg.m/s