Ashy A.
asked • 02/21/21physics question
A 0.40 kg model airplane is traveling 20 km/h toward the south. A 0.50 kg model airplane traveling 25 km/h in a direction 20° east of south collides with the first model airplane. The two planes stick together on impact. What is the direction and magnitude of the velocity of the combined wreckage immediately after the collision?
(answer is 22km/h and 12o East of south and i'm not too sure how to get to these answers)
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1 Expert Answer
Anthony T. answered • 02/21/21
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The fact that the planes stick together indicates that the collision is inelastic. Conservation of momentum applies, so the total momentum before collision equals the total momentum after collision.
Assume an x,y coordinate system. The plane heading south would correspond to the negative y direction. This plane initially would have only a y component of momentum, M1y = - 0.4 kg x 20kph. the second plane will have two momentum components, M2x = 0.5kg x 25kph x cos 20°, and M2y = - 0.5kg x 25kph x cos20°.
The momentum after collision will have a mass equal to the sum of the two plane's masses, 0.5 + 0.4 = 0.9kg.
This momentum will also have x and y components equal to 0.9kg x Vcx and -0.9kg x Vcy where Vc is the final velocity of the combined mass.
Now, set the x and y components before collision equal to to the x, y components after collision.
0.9 kg x Vcx = 0.5kg x 25kph x sin 20° and 0.9 kg x Vcy = - 0.4 kg x 20kph - 0.5kg x 25kph x cos20°
Solve these equations for Vcx and Vcy.
Vcx = 0.5kg x 25kph x sin 20° / 0.9kg = 4.75 kph
Vcy = - 0.4 kg x 20kph - 0.5kg x 25kph x cos20° / 0.9kg = -21.9 kph
The magnitude of the final velocity is √ 4.752 + (-21.9)2 = 22 kph
The direction is arctan -21.9 / 4.75 = -77.8° which corresponds to and angle of 90 - 77.8 = 12.2° east of south.
Draw a diagram which will better help to visualize the problem.
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George W.
Hello Ashy! Thanks so much for posting the question. These types of questions remind me of how much I need to review ( lol ). I believe I've stumbled upon the correct answer. If that isn't the case, please let me know. In this circumstance, the momentum possessed by each model airplane will combine after the collision in an inelastic collision. Recall the momentum ( linear ) can be summed up with p = ( mv ), where m = mass in kilograms, and v = velocity ( typically in meters per second ). Therefore, the following equation will be of interest to us: ( p1 initial ) + ( p2 initial ) = ( p1,2 final ) Momentum is sometimes referred to as a " quantity of motion ". In order to determine the " quantity of motion " that exists in the system before the crash, we must add the ( x ) and ( y ) components of p1 and p2. 1. p1 only has a ( y ) component ( South ), so ( ( 0.40 kg )( 20 km / h ) ) = ( 8 kg*km / h ). 2a. p2 has ( x ) and ( y ) components of motion relative to a 20-degree angle ( theta ). Recall that sin ( theta ) = ( opp / hype ). Here, ( opp ) = the ( x ) component of motion, and the ( hype ) = ( ( 0.50 kg )( 25 km / h ). As a consequence, ( x ) = ( ( 0.50 kg )( 25 km / h )( sin 20 degrees ) ) = ( 4.3 kg*km / h ). 2b. The ( y ) component of p2 is found using cos ( theta ) = ( adj / hype ). Therefore, ( y ) = ( ( 0.50 kg )( 25 km / h )( cos 20 degrees ) ) = ( 12 kg*km / h ). 3. The NET motion that exists between these two model aircraft is found using the Pythagorean Theorem. If ( x^2 ) + ( y^2 ) = ( p initial )^2, the net momentum of the system before the collision is ( p initial ) = sqrt ( x^2 + y^2 ), where ( x ) = ( 4.3 kg*km / h ), and ( y ) = ( ( 8 kg*km / h ) + ( 12kg*km / h ) ). 4. The NET momentum of the system before collision is ( p initial ) = sqrt [ ( 4.3 kg*km / h )^2 ) + ( ( 20 kg*km / h )^2 ) ]. ( p initial ) = ( 20 kg*km / h ). 5. ( p initial ) = ( p final ). So ( 20 kg*km / h ) = ( p final ). Recall that ( p ) = ( mv ). In this case, the final mass is the combined mass of the two planes. Therefore, ( 20 kg*km / h ) = ( 0.9 kg )( ? km / h ). Dividing both sides of the equation by the combined mass gives us the final velocity of ( 22 km / h ). 6. After the collision occurs, the combined masses will have adapted the ( x ) and ( y ) components of motion that occurred before the collision. Since tan ( theta ) = ( x / y ), the tan^-1 ( x / y ) = ( theta ) = tan^-1 [ ( 4.3 kg*km / h ) / ( 20 kg*km / h ) ] = 12 degrees East of South. The final answer ( should be ) 12 degrees East of South.02/21/21