points SenCP10 6.AE.004. EXAMPLE 6.4 A Truck versus a Compact GoAL Apply conserv
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points SenCP10 6.AE.004. EXAMPLE 6.4 A Truck versus a Compact GoAL Apply conservation of momentum to a one-dimensional nelastic collision PROBLEM A pickup truck with mass 1.8o x 103 kg is traveling eastbound at +15.0 m/s, while a compact car with mass 9.00 x 10 kg is traveling westbound at -15.0 m/s. (See figure.) The vehicles collide head-on, becoming entangled. (a) Find the speed of the entangled vehicles after the collision. (b) Find the change in the velocity of each vehicle. (c) Find the change in the kinetic energy of the system consisting of both vehicles. STRATEGY The total momentum of the vehicles before the collision, p, equals the total momentum of the vehicles after the collision, py if we ignore friction and assume the two vehicles form an isolated ystem. (This is called the "impulse approximation. solve the momentum conservation equation for the final velocity of the vehicles Once the velocities are in hand, the other parts can be solved by entangled substitution SOLUTION (A) Find the final speed after collision. Let ms and Viv represent the mass (mi m2 and initial velocity of the pickup truck, while m2 and va pertain to the compact. Apply conservation of momentum. (1.80 x 103 kg)(15.0 m/s) (9.00 x 102 kg)(-15.0 m/s) Substitute the values and solve for the (1.80 x 10 kg 9.00 x 102 kg)vr +5.00 m/s (B) Find the change in velocity for each vehicle. ava vr vs 5.00 m/s 15.0 m/s 10.0 m/s Change in velocity of the pickup truck. ava vi va 5.00 m/s (-15.0 m/s 20.0 m/s Change in velocity of the compact car. (c) Find the change in kinetic energy of the system. Calculate the initial kinetic energy of 3/2/2017 http://www.webassign.net/webstudent/Assignment-Responses last?dep 15377378 http://www.webassign neuweuw3www.iwi ....e..........Explanation / Answer
EXERCISE:
(A) Applying momentum conservation for the collision,
(9.14 x 10^2) (5.91) + (1.76 x 10^3) (18.4) = (914 + 1760) v
v= 14.13 m/s
(B) for trcuk: deltaV = 14.13 - 18.4 = - 4.27 m/s
for car: = 14.13 - 5.91 = 8.22 m/s
(C) Ki = (914 x 5.91^2 /2 ) + (1760 x 18.4^2 / 2) = 313894.94 J
Kf = (914 + 1760) (14.13^2) /2 = 266941.28 J
change in KE = Ki - Kf = 46953.7 J
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