A 0.149 kg glider is moving to the right on a frictionless, horizontal air track

Question

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.790 m/s . It has a head-on collision with a 0.310kg glider that is moving to the left with a speed of 2.10 m/s . Suppose the collision is elastic.

Part A

Find the magnitude of the final velocity of the 0.149 kg glider.

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Part B

Find the direction of the final velocity of the 0.149 kg glider.

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Part C

Find the magnitude of the final velocity of the 0.310 kg glider.

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Part D

Find the direction of the final velocity of the 0.310 kg glider.

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.790 m/s . It has a head-on collision with a 0.310kg glider that is moving to the left with a speed of 2.10 m/s . Suppose the collision is elastic.

Part A

Find the magnitude of the final velocity of the 0.149 kg glider.

m/s

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Part B

Find the direction of the final velocity of the 0.149 kg glider.

Find the direction of the final velocity of the 0.149  glider. to the right to the left

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Part C

Find the magnitude of the final velocity of the 0.310 kg glider.

m/s

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Part D

Find the direction of the final velocity of the 0.310 kg glider.

Find the direction of the final velocity of the 0.310  glider. to the right to the left

Explanation / Answer

m1 = 0.149 kg   m2 = 0.31 kg

speeds before collision

u1 = 0.79 m/s     u2 = -2.1 m/s

speeds after collision

v1 = ?        v2 = ?

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from moentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2

(0.149*0.79) - (0.31*2.1) = (0.149*v1) + (0.31*v2).....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

0.5*0.149*0.79^2 + 0.5*0.31*2.1^2 = 0.5*0.149*v1^2 + 0.5*0.31*v2^2 ........(2)

v2 = 20.9 m/s <---------------answer

solving 1 & 2

v1 = -3.114 m/s

v2 = -0.224 m/s

part(A)

v1 = 3.114 m/s

part(B)

left

part(C)

v2 = 0.224 m/s

part(D)

left

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