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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 9-65). At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface The collision is elastic. Find
(a)the speed of the ball and (b) the speed of the block, both just after the collision.


Sagot :

(a) ball velocity = - 2.47 m/s  

(b) block velocity = 1.236 m/s

What is collision?

Two items collide when they briefly come into touch with one another. To put it another way, a collision is a brief reciprocal encounter between two masses in which the momentum and energy of the masses change.

For ball as it is at initially at height of 70.0 cm = 0.7 m

so by energy conservation,

K.E.(i) P.E.(i) + K.E.(f)+P.E.(f)

0 + mg×(0.7)  = 1/2 mv² + 0

v = 3.71 m/s

now, collision b/w ball and block

as the block is in rest initially and there is no external force b/w them

so by momentum conservation,

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂

or, (0.5 x 3.71) + 0 = (0.5x v₁) + (2.5 x v₂)

3.71 = v₁ + 5v₂  .........(1)

as elastic collision so energy conservation

or, 1/2 m₁ u₁² + 1/2 m₂ u₂² = 1/2 m₁ v₁² + 1/2 m₂ v₂²

or, (1/2 x 0.5 x 3.712) + 0 = (1/2 x 0.5 x v₁²) + (1/2 x 2.5 x v₂²)

or, 3.433 = 0.25xv₁² + 1.25 v₂²      ...............(2)

from eq. 1 and eq. 2

or, 3.433 = 0.25 x (3.71 - 5 v₂)² + 1.25v₂²  

or, v² = 1.236 m/s

and v₁ = 3.71 - 5v₂

or, v₁ = - 2.47 m/s

so after collision

ball velocity = - 2.47 m/s       [(-)ve x direction]

block velocity = 1.236 m/s     [(+)ve x direction]

To know more about velocity refer to:

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