Twochildren are playing on áform lake by throwing di???rent balk at a mall ice b

Question

Twochildren are playing on áform lake by throwing di???rent balk at a mall ice block. itting on top of the ice surface. The ice block han a mass of 10.okg The children throw a baskethall, and a tennis ball at the initially stationary block Then they slide the ice block tomards the stationnry balls (I) (10 points) Ir the children throw the 0625kg-bakteball at ?OOmj. find the a locity of ball and block after the collision (Assume elastic.) (a) (10 points) If the children throw the 58.5g-tennis ball at 10.0/s find the velocity of ball and block after the collision.(Assume elastic.) (II) (10 points) If the ice-block was thrown at 1.50 "/s, what speed each of the two balls have after the collision, if they started at rest? (Assume elastic.)

Explanation / Answer

Let the mass of the ice block be M = 10.0 kg

Let the mass of the basket ball is m1 = 0.625 kg

Let the mass of the tennis ball be m2 = 58.5 g = 0.0585 kg

In a head on collision between two masses m1 and m2 moving with speeds u1 and u2 , final speeds are given by

v1 = ( 2 m2 u2 / m1 + m2 ) + { ( m1 - m2 ) u1 / m1 + m2 } ....... final speed of mass m1

v2 = ( 2m1 u1 / m1 + m2 ) + { ( m2 - m1 ) u2 / m1 + m2 } ........ final speed of the mass m2

If the second mass is at rest ,u2 = 0 . Then above eqns reduce to

v1 = { ( m1 - m2 ) u1 / m1 + m2 }

and v2 = ( 2 m1 u1 / m1 + m2 )

A) Ice block is at rest , so it ' speed is zero .

Initial speed of the basket ball be u1 = 10 m/s

Final speed of the basket ball be v1 and ice block be V after collision

v1 = ( m1 - M ) u1 / m1 + M

= { ( 0.625 - 10 ) × 10 } / ( 0.625 + 10 )

= - 9.375/10.625

= - 8.82 m/s

Ice block is too heavy and it reverses the direction of the basketball. So the speed of the basketball is negative.

Final speed of the ice block is calculated as follows

V = 2 m1 u1 / m1 + M

= ( 2 × 0.625 × 10 ) / ( 10 + 0.625 )

= 12.50/10.625

= 1.176 m/s

Speed of the ice block is too less because it is much heavier than basketball and so it's impact on ice block is very less.

B) Let the initial speed of the tennis ball be u2 = 10 m/s

Final speed of the tennis ball be v2 and ice block be V' after collision.Then

v2 = ( m2 - M ) u2 / m2 + M

= ( 0.0585 - 10 ) × 10 / ( 0.0585 + 10 )

= - 99.415/10.0585

= - 9.88 m/s

V' = ( 2 × m2 × u2 ) /( m2 + M )

= ( 2 × 0.0585 × 10 )/ ( 0.0585 + 10 )

= 1.17 / 10.0585

= 0.116 m/s

This can be taken as zero for practical purposes.

Actually mass m2 of tennis ball is very much less than mass of the ice block M.

In such cases we can take m2 <<< M so that speed of the tennis ball becomes v2 = -10 m/s and speed of the ice block is V' = 0

C) Consider the collision between the moving ice block and basketball at rest.

Take the speed of the ice block as U = 1.5 m/s

Initial speed of the basketball be U' = 0

Final speed of the ice block be v' and basketball be V1

Then , v' = ( M - m1 ) U / ( M + m )

= { ( 10 - 0.625 ) × 1.5 } / ( 10 + 0.625 )

= 1.323 m/s

So the speed of the ice block decreases after collision with the basketball.

V1 = 2 M U / ( M + m1 )

= ( 2 × 10 × 1.5 ) / ( 10 + 0.625 )

= 30 / 10.625

= 2.82 m/s

So the speed of the basketball is V1 = 2.82 m/s

Now consider the collision between the ice block and tennis ball at rest.

Let the final speed of the ice block is v'' and tennis ball is V2 . Then

v'' = ( M - m2 ) U / ( M + m2)

= { ( 10 - 0.0585 ) × 1.5 } / ( 10 + 0.0585 )

= 14.91 / 10.0585

= 1.482 m/s

Since the mass of the tennis ball is too less , speed of the ice block has not changed much.

V2 = 2 M U / ( m1 + m2 )

= ( 2 × 10 × 1.5 ) / ( 10 + 0.0585 )

= 30 / 10.0585

= 2.982 m/s

So the final speed of the tennis ball is V2 = 2.982 m/s

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