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A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.8 kg, is moving upward at 22 m/s and the other ball, of mass 2.1 kg, is moving downward at 12 m/s. How high do the combined two balls of putty rise above the collision point

Sagot :

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Answer:

the balls reached a height of 4.9985 m

Explanation:

Given the data in the question;

mass one m = 3.8 kg

mass two M = 2.1 kg

Initial velocities

u = 22 m/s

U = { moving downward} = 12 m/s

Now, using the law conservation of linear moment;

mu + MU = v( m + M )

we solve for "v" which is the velocity of the ball s after collision;

v = (mu + MU) / ( m + M )

so we substitute our given values into the equation

v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )

v = ( 83.6 - 25.2 ) / 5.9

v = 58.4 / 5.9

v = 9.898 m/s

Now, we determine required height using the following relation;

v"² - v² = 2gh

where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²

0 - v² = 2gh

v² = -2gh

so we substitute

( 9.898 )² = -2 × -9.8  × h

97.97 = 19.6 × h

h = 97.97 / 19.6

h = 4.9985 m

Therefore, the balls reached a height of 4.9985 m

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