Steel Ball and Block A steel ball of mass 0.500 kg is fastened to a cord that is

Question

Steel Ball and Block A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 10-61). At the bottom of its path, the ball strikes a 2.00 kg steel block initially at rest on a frictionless surface. The collision is elastic.

Figure 10-61
(a) Find the speed of the ball just after collision.
m/s
(b) Find the speed of the block just after collision.
m/s


DIAGRAM

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Explanation / Answer

a .

speed of ball at bottom = v = ( 2 g h ) = ( 2 *9.8 * 0.7 ) = 3.7040 m /s ;

momentum conservation :

0.5 * 3.7040 = 0.5 v1 + 2 v2 ; v2 = ( 1.852 - 0.5 v1 ) /2 = ( 0.926 - 0.25 v1 );

since KE is conserved ,

( 1/2 ) 0.5 * 3.7040^2 = [ 1 /2 ][ 0.5 v1^2 + 2 v2^2 ] ;

or   0.5 * 3.7040^2 = [ 0.5 v1^2 + 2 v2^2 ] ;

since v2 = ( 0.926 - 0.25 v1 )

or   0.5 * 3.7040^2 = [ 0.5 v1^2 + 2 ( 0.926 - 0.25 v1 )^2 ] ;

or 6.859 = 0.5 v1 ^2 + 2 ( 0.857 + 0.0625 v1^2 - 0.463 v1 ) ;

or v1 = 3.7040 , -2.22 ; we are getting two values , one of them is correct

v2 = 0 , 1.481 ,

speed of ball after collison = 3.7040 OR -2.22 ;

speed of block after colluision = v2 = 0 or 1.481

obviously , v2 = 0 and v1 = 3.7040 are not possible in a elastic collsioin ,

so the correct answers are

speed of ball after collison = -2.22 m /s

speed of block after colluision = 1.481 m /s

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