Question

A frictionless steel ball of radius 2cm, moving on a
horizontal plane with a velocity of 5cm/s, collides
head-on with another stationary steel ball of radius
3cm. The velocities of two bodies after collision will
respectively be (in cm/s) (e = 1) :-
(1) 2.7, 2:3
(2) -2.7, 2-3
(3) 2.7,-2.3
(4) -2.7, -2.3

Answer 1

Answer:

2nd option

Explanation:

please refer to attachment!!

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Answer 2

Answer:

The velocities of two bodies after collision is -2.7 cm/s and 2.3 cm/s.

(2) is correct option.

Explanation:

Given that,

Radius = 2 cm

Velocity = 5 cm/s

Radius of another ball = 3 cm

Here, both balls of steel so same density.

The mass will be proportional to the volume and volume is proportional to  cube of radius.

So, the masses of balls will be

$m_{1}=8m$

$m_{2}=27m$

We need to calculate the velocities of two bodies after collision

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$....(I)

Using newton's law of collision

$\dfrac{u_{1}-u_{2}}{v_{2}-v_{1}}=e$

Put the value into the formula

$\dfrac{5-0}{v_{2}-v_{1}}=1$

$v_{2}-v_{1}=5$

$v_{2}=v_{1}+5$...(II)

Put the value of v₂ in equation (I)

$8m\times5+0=8mv_{1}+27m(v_{1}+5)$

$40=8v_{1}+27v_{1}+135$

$35v_{1}=-95$

$v_{1}=\dfrac{-95}{35}$

$v_{1}=-2.7\ cm/s$

Put the value of v₁ in equation of (II)

$v_{2}=-2.7+5$

$v_{2}=2.3\ cm/s$

Hence, The velocities of two bodies after collision is -2.7 cm/s and 2.3 cm/s.