Problem 9.34 A 0.060-kg tennis ball, moving with a speed of 4.3 m/s , has a head

Question

Problem 9.34

A 0.060-kg tennis ball, moving with a speed of 4.3 m/s , has a head-on collision with a 0.10-kg ball initially moving in the same direction at a speed of 2.6 m/s .

Part A

Assuming a perfectly elastic collision, determine the speed of each ball after the collision.

Enter your answers numerically separated by a comma. Express your answers using two significant figures.

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Part B

Determine the direction of tennis ball after the collision.

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Part C

Determine the direction of 0.10-kg ball after the collision.

Problem 9.34

A 0.060-kg tennis ball, moving with a speed of 4.3 m/s , has a head-on collision with a 0.10-kg ball initially moving in the same direction at a speed of 2.6 m/s .

Part A

Assuming a perfectly elastic collision, determine the speed of each ball after the collision.

Enter your answers numerically separated by a comma. Express your answers using two significant figures.

vtennis ball, vball =   m/s  

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Part B

Determine the direction of tennis ball after the collision.

Determine the direction of tennis ball after the collision. The tennis ball moves in the direction of its initial motion. The tennis ball moves in opposite direction.

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Part C

Determine the direction of 0.10-kg ball after the collision.

Determine the direction of 0.10- ball after the collision. The ball moves in the direction of its initial motion. The ball moves in opposite direction.

Explanation / Answer

m1(tennis) = 0.06 kg                    m2(ball) = 0.1 kg

speeds before collision

v1i = 4.3 m/s                   v2i = 2.6 m/s

speeds after collision

v1f = ?                         v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ( (0.06-0.1)*4.3 + (2*0.1*2.6))/(0.06+0.1)

V1f = 2.175 m/s

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f =   ( (0.1-0.06)*2.6 + (2*0.06*4.3))/(0.06+0.1)

v2f = 3.875 m/s

part(A)

Vtennisball = 2.2 m/s

Vball (v2f) = 3.9 m/s

part(B)

The tennis ball moves in the direction of its initial motion.

part(D)

The ball moves in the direction of its initial motion.

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