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3 years ago

Which one? A? B? C? D?

Physics

1 answer:

lidiya [134]3 years ago

5 0

Answer:The answer is D.

Explanation:

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One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

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

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

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0

3 years ago