Respuesta :
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
[tex]m = m_{1}=m_{2}[/tex]
(a). We need to calculate the speed of the first ball after the collision
Using law of conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
Along X- axis
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}[/tex]
[tex]v_{1}=u_{1}+u_{2}[/tex]
Put the value into the formula
[tex]v_{1}=2.2i-0.80i[/tex]
[tex]v_{1}=1.4i\ m/s[/tex]
Along Y-axis
[tex]0=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]m_{1}v_{1}=-m_{2}v_{2}[/tex]
[tex]v_{1}=-v_{2}[/tex]
Put the value into the formula
[tex]v_{1}=-1.36j\ m/s[/tex]
Then the final speed of the first ball
[tex]v_{1}=\sqrt{(1.4)^2+(1.36)^2}[/tex]
[tex]v_{1}=1.95\ m/s[/tex]
(b) We need to calculate the direction of the first ball after the collision
Using formula of direction
[tex]\tan\theta=\dfrac{v_{2}}{v_{1}}[/tex]
[tex]\tan\theta=\dfrac{-1.36}{1.4}[/tex]
[tex]\theta=\tan^{-1}\dfrac{-1.36}{1.4}[/tex]
[tex]\theta=-44.16^{\circ}[/tex]
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.
(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.
What is the law of conservation of momentum?
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The momentum of the body is given by the product of the mass and velocity of the body.
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
(a). The speed of the first ball after the collision is 1.95 m/s
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
[tex]\rm m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
Along the x-axis, the principle of momentum conservation is applied
[tex]\rm m_1u_1+m_2u_2=m_1v_1[/tex]
[tex]\rm v_1= u_1+u_2[/tex]
[tex]\rm v_1= 2.2 \vec i-0.80 \vec i \\\\ \rm v_1=1.41 \ \vec i m/sec[/tex]
Along the y-axis, the principle of momentum conservation is applied
[tex]\rm m_1v_1+m_2v_2= 0 \\\\\ \rm m_1v_=-+m_2v_2 \\\\ v_1=-v_2 \\\\ v_1= -1.36 \vec j \ m/sec[/tex]
The net speed of the first ball is found as;
[tex]\rm v_1= \sqrt{(1.4)^2+(1.36)^2} \\\\ \rm v_1=1.95 \ m/sec[/tex]
(b). The direction of the first ball after the collision is 44.16° due south of east.
To find the direction tangent formula is used;
[tex]\rm tan \theta = \frac{v_2}{v_1} \\\\ \rm tan \theta = \frac{-1.36}{1.4} \\\\ \theta= tan^{-1 } \frac{-1.36}{1.4} \\\\ \theta= -44.16^0[/tex]
Hence the direction of the first ball after the collision is 44.16° due south of east.
To learn more about the law of conservation of momentum refer to;
https://brainly.com/question/1113396
