A 2-kg ball moving to the right with a velocity of 5 m/s strikes at A the surfac

Question

A 2-kg ball moving to the right with a velocity of 5 m/s strikes at A the surface of a 10-kg quarter cylinder which is initially at rest and in contact with a spring of constant 10 kN/m. The spring is held by cables so that it is initially compressed 50mm. Neglecting friction and knowing that the coefficient of restitution is 0.7, determine (a) the velocity of the ball immediately after impact, (b) the maximum distance the quarter cylinder moves, and (c) the impulse the ball exerts on the quarter cylinder during impact. (All related diagrams required for full credit.)

Explanation / Answer

1.Considering momentum

So,

equation becomes

m1v1+m2v2=m1v1'+m2v2'

v2=0 m/s

v1=5*0.7=3.5m/s

So,

10=2(v2'-35)+10v2'

Since,

v2'-v1'/v1-v2

SO

v2'=6.67m/s

v1'=28.33 m/s

B)Maximum distance= 1/2m2(v2cos40)^2

Substituting values,

We get

distance = 5.05m

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