Problem 6.113 Part A Ignoring friction, what\'s the speed of the joined cars at

Question

Problem 6.113 Part A Ignoring friction, what's the speed of the joined cars at the bottom of the incline? A 980 kg car is at the top of a 54 m -long, 25° incline. Its parking brake fails and it starts rolling down the hill. Halfway down, it strikes and sticks to a 1190 kg parked car. Express your answer to two significant figures and include the appropriate units. H n o . , v= Value Units Submit My Answers Give Up . Part B What the first car's speed would have been at the bottom had it not struck the second car Express your answer to two significant figures and include the appropriate units. He hn O C o ? v= Value | Units Submit my answers give up

Explanation / Answer

Given

m1 = 980 kg , m2 = 1190 kg

length of the ramp at an inclination theta 2.5 degrees is s= 54 m

hlaf the way down is at 27 m , strikes m2 which is at rest

Part a

the speed of m1 beforw it strikes the m2 is v^2 - u^2 = 2as

v^2 -0 = 2g*s sin theta

v = sqrt(2g*s sin theta )

v = sqrt(2*9.8*27 sin2.5) m/s = 4.806 m/s

now the conservation of momentum is m1u1+m2u2 = (m1+m2)v

980*4.806 + 1190*0 = (980+1190)*V

v = 2.170 m/s

both are movign with speed v = 2.170 m/s from 27 m point and reaches the bottom so the speed of the two cars at bottom is  

v = sqrt(2g*s sin theta )

v= sqrt(2*9.8*27 sin 2.5) = 4.806 m/s

part b

the car m1 has speed at the bottom if it was not stuck with m2 is  

by conservation of energy  

gravitational potential energy = kinetic energy

m1g sin theta*s = 0.5*m1v^2

v^2 = 2*g sin theta*s

v = sqrt(2*g sin theta*s)

v = sqrt(2*9.8 sin 2.5*54) = 6.8 m/s

the speed of first carn when it was not stuck with m2 is v = 6.8 m/s

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