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Shtirlitz [24]
3 years ago
8

A 0.060-kg tennis ball, moving with a speed of 5.50m/s , has a head-on collision with a 0.090-kg ball initially moving in the sa

me direction at a speed of 3.40m/s . Assume that the collision is perfectly elastic.
A. Determine the speed of the 0.060-kg ball after the collision.
B. Determine the direction of the velocity of the 0.060-kg ball after the collision.
Determine the direction of the velocity of the 0.060- ball after the collision.
1. in the direction of the initial velocity
2. in the direction opposite to the initial velocity
C. Determine the speed of the 0.090-kg ball after the collision.
D. Determine the speed of the 0.090-kg ball after the collision.
Determine the direction of the velocity of the 0.090- ball after the collision.
1. in the direction of the initial velocity
2. in the direction opposite to the initial velocity
Physics
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

Explanation:

For elastic collision , the formula for velocity of .06 kg  is

v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂

=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4

=( -.03 / .15) x 5.5 + (2 x .0054 / .15) x 3.4

= -1.1 +.2448

= - 0 .8552 m/s

Its direction will be - opposite direction

the formula for velocity of .09kg  is

v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁

= ( .09-.06 / .06+.09 ) 3.4 + (2 x .06 x .09 / .06+.09 ) 5.5

= .68 +.396

= 1.076 m / s

Its direction will be in the same direction .

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Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

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