Two bumper cars are sliding toward one another. initially, car 1 is traveling to

Question

Two bumper cars are sliding toward one another. initially, car 1 is traveling to the east at 5.62 m/s, and car 2 is traveling 60.0 degrees south of West at 10.0 m/s. After they collide, bumper car 1 is observed go be traveling to the West with a speed of 3.14 m/s. Friction is negligible between the cars and the ground.

a) if the masses of cars 1 and 2 are 596 kg and 625 kg, what is the velocity of car 2 immediately after the collision?

b) what is the kinetic energy lost in the collision?

s traveling to the Easta5.62 m/s, and lide, bumper car 1 is obscrved to be ble between the cars and the ground. g, respectively, what is the velocity of car

Explanation / Answer

here,

initial velocity , u1 = 5.62 m/s i

u2 = 10 * ( - cos( 60) i - sin(60) j )

u2 = - 5 m/s i - 8.66 m/s j

m1 = 596 kg

m2 = 625 kg

final velocity , v1 = - 3.14 j

let the final velocity of car 2 be ( v2x i + v2y j)

using conservation of mmomentumalong y axis

- m2*8.66 = m2*v2y

v2y = - 8.66 m/s

using conservation of momentum along x axis

m1*5.62 - m2 * 5 = - m1* 3.14 + m2*v2x

596*5.62 - 625 *5 = - 596 * 3.14 + 625 *v2x

v2x = 3.35 m/s

v2 = ( 3.35^2 + 8.66^2)

v2 = 9.29 m/s

a)

the velocity of car 2 immediately after the collison is 9.29 m/s

b)

the kinetic energy lost in the collison , KE = 0.5 *( m1*u1^2 + m2*u2^2 - m1*v1^2 - m2*v2^2)

KE = 0.5 * ( 596 * 5.62^2 + 625 * 10^2 - 596 * 3.114^2 - 625*9.29^2)

KE = 1.08 * 10^4 J

the kinetic energy lost in the collison is 1.08 * 10^4 J

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