please do parts a&b Class Management Help Homework #il (Ch 9: Momentum, In pulse
ID: 1796005 • Letter: P
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please do parts a&b
Class Management Help Homework #il (Ch 9: Momentum, In pulse and Collisio Begin Date: 916201 I 200:00 12/20/2017 12.00:00 AM M-Due Date: 12,S 201 T -so ost E d ille (13%) Problem 8: A car with mass me-1014 kg is traveling west through an intersection at a magnitude of velocity. ofve-6.5 m/s when a truck of mass mt = 1889 kg traveling south at ½ = 7.2 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of No) friction of 0.5. Status e for Efx) theexpertta c Status ompleted omp ompleted ompleted 50% Part (a) Write an expression for the velocity ofthe system after the collision, in terms of the ranables man the problem aat met and the unit vectors i and j 50% Part (b) How far, in meters, will the vehicles slade after the collision, Grade Summary Deductions Poteatial100 artial Subaissions A ttempts reosining cotan0 asinOacos0 atan0 acotan) sinh) pet attemp) detaledt view * Degrees RadiansExplanation / Answer
a) Apply conservation of momentum in x-direction
mc*(-vc) = (mc + mt)*vx
==> vx = -mc*vc/(mc +mt)
Apply consrvation of momentum in y-direction
mt*(-vt) = (mc + mt)*vy
==> vx = -mt*vt/(mc +mt)
velocity of the system after the collsion,
v = vx i + vy j
= -mc*vc/(mc +mt) i - mt*vt/(mc + mt) j
b) vx = -1014*6.5/(1014 + 1889)
= -2.27 m/s
vy = -1889*7.2/(1014 + 1889)
= -4.68 m/s
v = sqrt(vx^2 +vy^2)
= sqrt(2.27^2 + 4.68^2)
= 5.20 m/s
accelration of the vehicles, a = -g*mue_k
= -9.8*0.5
= -4.9 m/s^2
distance travelled before stopping,
d = (vf^2 - vi^2)/(2*a)
= (0^2 - 5.2^2)/(2*(-4.9))
= 2.76 m <<<<<<<<<<------------------------Answer
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