Aball 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. 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, 1.67 kg and 2.59 kg, and the length of the wire is 1.01 m. find the velocity of the ball just after the collision.

0.96 m/s

Explanation:

Consider the motion of the ball before collision with the block

h = height from which the ball is dropped = length of the wire = 1.01 m

m = mass of the ball = 1.67 kg

v = speed of the ball just before collision with the block = ?

Using conservation of energy for the ball

Kinetic energy of the ball at the bottom = Potential energy of the ball at the top

(0.5) m v² = mgh

(0.5) v² = gh

(0.5) v² = (9.8) (1.01)

v = 4.45 m/s

consider the collision between the ball and the block :

m = mass of the ball = 1.67 kg

v = velocity of the ball just before collision with the block = 4.45 m/s

v' = velocity of the ball just after collision with the block

M = mass of the block = 2.59 kg

V = velocity of the block just before collision with the ball = 0 m/s

V' = velocity of the block just after collision with the ball

Using conservation of momentum

mv + MV = mv' + MV'

(1.67) (4.45) + (2.59) (0) = 1.67 v' + (2.59) V'

7.43 = 1.67 v' + (2.59) V'

                               eq-1

Using conservation of kinetic energy

(0.5) mv² + (0.5) MV² = (0.5) mv'² + (0.5) MV'²

mv² + MV² = mv'² + MV'²

(1.67) (4.45)² + (2.59) (0)² = 1.67 v'² + (2.59) V'²

using eq-1

(1.67) (4.45)²  = 1.67 v'² + (2.59) ((7.43 - 1.67 v')/2.59)²

v' = - 0.96 m/s

the negative sign indicates the direction which is opposite to its direction before colliding with the block.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j