A ball is attached to one end of a wire, the other end being fastened to the cei

Question

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.5 kg and 2.5 kg, and the length of the wire is 1.36 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b)just after the collision.

Explanation / Answer

1.)

Let the velocity of the ball before collision be u

Thus applying the conservation of Mechanical Energy=PE+KE

ME at starting=1.5g(1.36)=ME before collision=1.5(u^2)

Thus u=sqrt(g(1.36))=sqrt((9.8)(1.36))=sqrt(13.328)=3.651 m/s

The direction would only be in +x axis as velocity of the ball is always tangential to the rope.

Thus, u=3.651 i m/sec where i is the unit vector in x axis

2.)

In elastic collision-

a.) Velocity of approach= Velocity of seperation

Let the velocity of the ball after the collision be v1 and velocity of the block after the collision be v2

Thus 3.651 = v1-v2 .............(Equ1)

And,

b.) Energy is conserved. Thus,

1.5g(1.36)= 1.5(v1^2)/2+2.5(v2^2)/2

39.984= 3(v1^2)+5(v2^2) .............(Equ2)

Solving (Equ1) and (Equ2)

(v1,v2)=(0.9131,-2.74) or (3.650625,-0.000275)

Thus v1=0.9131 i or 3.650625 i

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