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Sagot :
Answer:
[tex]14.98\ \text{kg m/s}[/tex]
[tex]45.26^{\circ}[/tex]
Explanation:
[tex]P_1[/tex] = Initial momentum of the pin = 13 kg m/s
[tex]P_i[/tex] = Initial momentum of the ball = 18 kg m/s
[tex]P_2[/tex] = Momentum of the ball after hit
[tex]55^{\circ}[/tex] = Angle ball makes with the horizontal after hitting the pin
[tex]\theta[/tex] = Angle the pin makes with the horizotal after getting hit by the ball
Momentum in the x direction
[tex]P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}[/tex]
Momentum in the y direction
[tex]P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}[/tex]
[tex](P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}[/tex]
The pin's resultant velocity is [tex]14.98\ \text{kg m/s}[/tex]
[tex]P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}[/tex]
The pin's resultant direction is [tex]45.26^{\circ}[/tex] below the horizontal or to the right.
The megnitude and the direction of the resultant velocity of the pin willl be 14.98 Kgm/sec and 45.26° below the horizontal or to the right.
What is momentum?
The momentum is defined as the product of mass and the velocity of the body. It is denoted by the letter P. It occurs due to the applied force. Its unit is Kg m/s².
The given data in the problem is;
P₁ is the initial momentum of the pin = 13 kg m/s
[tex]\rm p_i[/tex] is the initial momentum of the ball = 18 kg m/s
P₂ is the momentum of the ball after hit
[tex]\rm \theta[/tex] is the angle the pin makes with the horizotal after getting hit by the ball
Momentum in x direction;
[tex]\rm P_i= P_1 cos \theta + P_2 cos \theta \\\\ P_2cos \theta = P_i -P_i cos 55^0 \\\\ P_2cos \theta = 10.54 \ Kgm/s[/tex]
Momentum in y direction;
[tex]\rm P_i sin\theta = P_2 sin\theta \\\\ P_2 sin\theta =13 sin 15^0 \\\\ P_2 sin\theta =10.64 \ kgm/sec[/tex]
[tex](P_2 cos \theta )^2 + (P_2 sin \theta)^2 = P_2^2 \\\\ \rm P_2 = \sqrt{(10.54)^} +(10.64)^2[/tex]
[tex]\rm P_2 = 14.98 \ kgm/sec[/tex]
The direction of the resulatnt velocity is find as;
[tex]P_2 sin\theta = 10.64 \\\\ \theta = sin^{-1 } \frac{10.64}{14.98} \\\\ \theta=45.26^0[/tex]
Hence the megnitude and the direction of the resultant velocity of the pin willl be 14.98 Kgm/sec and 45.26° below the horizontal or to the right.
To learn more about the momentum refer to the link;
https://brainly.com/question/4956182
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