Question

Cow 1: 400 kg and going 3 m/s. Cow 2: 500 kg and going 2 m/s Two cows collide with different masses and velocities and jet off 40 degrees to the right of original direction and 50 degrees to the left of original direction. Find the velocity of each cow.

Answer 1

Answer on Question #73297-Physics-Mechanics-Relativity

Cow 1: 400 kg and going 3 m/s. Cow 2: 500 kg and going 2 m/s. Two cows collide with different masses and velocities and jet off 40 degrees to the right of original direction and 50 degrees to the left of original direction. Find the velocity of each cow.

Solution

From the conservation of momentum in x-direction:

$$m _ {1} v _ {1} + m _ {2} v _ {2} = m _ {1} v _ {1} ^ {\prime} \cos 4 0 + m _ {2} v _ {2} ^ {\prime} \cos 5 0$$

From the conservation of momentum in y-direction:

$$0 = m _ {1} v _ {1} ^ {\prime} \sin 4 0 - m _ {2} v _ {2} ^ {\prime} \sin 5 0$$ $$v _ {2} ^ {\prime} = v _ {1} ^ {\prime} \frac {m _ {1} \sin 4 0}{m _ {2} \sin 5 0}$$

So,

$$m _ {1} v _ {1} + m _ {2} v _ {2} = m _ {1} v _ {1} ^ {\prime} \cos 4 0 + m _ {2} v _ {1} ^ {\prime} \frac {m _ {1} \sin 4 0}{m _ {2} \sin 5 0} \cos 5 0$$ $$v _ {1} ^ {\prime} = \frac {m _ {1} v _ {1} + m _ {2} v _ {2}}{m _ {1} \cos 4 0 + m _ {1} \cot 5 0 \sin 4 0}$$ $$v _ {1} ^ {\prime} = \frac {v _ {1} + \frac {m _ {2}}{m _ {1}} v _ {2}}{\cos 4 0 + \cot 5 0 \sin 4 0}$$ $$v _ {1} ^ {\prime} = \frac {3 + \frac {5}{4} 2}{\cos 4 0 + \cot 5 0 \sin 4 0} = 4. 2 \frac {m}{s}.$$ $$v _ {2} ^ {\prime} = \frac {3 + \frac {5}{4} 2}{\cos 4 0 + \cot 5 0 \sin 4 0} \frac {4 \sin 4 0}{5 \sin 5 0} = 2. 8 \frac {m}{s}.$$

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