Respuesta :
Answer:
a. [tex]v=3.11mls[/tex], [tex]29.4^{0}[/tex]
b. [tex]K.E =-697.8J[/tex]
Explanation:
To calculate the values in the question, a deep understanding of perfect inelastic collision is important.
When two bodies undergo inelastic collision, two important parameters must be well understood i.e
Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision
Kinetic energy: Kinetic energy is not conserved due to dissipative force.
a.To calculate the velocity, we first find the total momentum before collision
Momentum of player 1 [tex]p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\[/tex]
Momentum of player 2 [tex]p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\[/tex]
Hence the total momentum [tex]p_{12}=p_{1}+p_{2}\\[/tex]
Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate
Hence [tex]p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\[/tex]
[tex](95+90)v=475i+270j\\[/tex]
[tex]v=2.57i+1.45j\\[/tex]
solving for the resultant velocity, we have
[tex]v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls[/tex]
To calculate the direction of movement, we have
[tex]\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}[/tex]
b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was
[tex]K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\[/tex]
And the final kinetic energy after collision is
[tex]K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J[/tex]
The decrease in Kinetic energy is
[tex]K.E =K.E_{final}- K.E_{initial}=894.7-1592.5[/tex]
[tex]K.E =-697.8J[/tex]
The negative sign indicate a decrease in Kinetic energy
The velocity of the players just after the tackle [tex]V_{f} =4.027 \frac{m}{s}[/tex] and the change in kinetic energy will be [tex]KE= 92J[/tex]
What will be the Velocity and kinetic energy of a player after an inelastic collision?
Given information
Mass of player one [tex]m_{1} =95\ kg[/tex]
The velocity of player one [tex]V_{1} = 5 \dfrac{m}{s}[/tex]
Mass of player second =[tex]m_{2} =90\ kg[/tex]
The velocity of player second = [tex]V_{2} = 3 \dfrac{m}{s}[/tex]
The final velocity of both objects will be = [tex]V_{F}\\[/tex]
Now we know that the collision is perfectly inelastic so from the conservation of momentum we can follow the expression below
[tex]m_{1} V_{1} +m_{2} V_{2}= (m_{1} +m_{2})V_{F}[/tex]
[tex](95\times 5)+ (90\times 3)= (90+95)\times V_{F}[/tex]
So the resultant value of velocity after the collision
[tex]V_{F}=4.027 \ \frac{m}{s^{2} }[/tex]
Now the change in kinetic energy will be calculated as
Kinetic energy initial = [tex]\dfrac{1}{2} m_{1}V_{1}^{2}+ \dfrac{1}{2} m_{2}V_{2}^{2[/tex]
[tex]\dfrac{1}{2} \times 95\times 5\times5+ \dfrac{1}{2} \times 90\times 3\times3[/tex]
[tex]KE= 1592.5\ J[/tex]
Similarly final KE of the player
KE= [tex]\dfrac{1}{2} (m_{1} +m_{2} )V_{F} ^{2}[/tex]
[tex]KE_{F} = \dfrac{1}{2}(95+90)\times 4.027\times 4.027[/tex]
[tex]KE_{F} = 1500\ J[/tex]
So the change in KE will be
[tex]=1592-1590=92 \ J[/tex]
Thus the velocity of the players just after the tackle [tex]V_{f} =4.027 \frac{m}{s}[/tex] and the change in kinetic energy will be [tex]KE= 92J[/tex]
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