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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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a block measures 3.5 cm long 2.8 cm wide and 1.6 cm deep. the density of the block 2.5 g/cm. calculate the volume of the block.
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An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
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Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

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2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

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3) and 4)

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So, we can write two equations, one for the x-component, and other for the y-component, as follows:

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v = √(15.4)²+(12.9)² = 20.1 m/s

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6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

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