a Police officer at an accident scene find that upon collision two entangled car move at 3.4 m\s at 44 degrees north east they look at 1000kg car that moving 15m\s due west before collision what was the magnitude and direction of velocity of 1500kg car ( ignore friction)
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If I understand correctly:
After the collision, the entangled cars (total mass = 2500 kg) were moving at 3.4 m/s at an angle 44º north of east. That's a momentum of:
p = mv = (2500 kg)(3.4 m/s) = 8500 kg*m/s @ 44º north of east
Before we do anything else, let's go ahead and resolve that momentum into its east-west and north-south components:
east-west component: (8500 kg*m/s)cos44º = 6114 kg*m/s east
north-south component: (8500 kg*m/s)sin44º = 5905 kg*m/s north
So...these were the east-west and north-south components of the total momentum BEFORE the collision as well as AFTER the collision.
Now let's examine the cars individually pre-collision. Car 1 (1000 kg) is moving due west at 15 m/s, which gives it a momentum of 15 000 kg*m/s west. It has no north-south momentum. For the sake of simplicity, let's call Car 1's pre-momentum collision -15 000 kg*m/s and the total east-west momentum +6114 kg*m/s.
p1 + p2 = ptotal
(-15,000 kg*m/s) + p2 = +6114 kg*m/s
p2 = 21 114 kg*m/s east
So that's Car 2's eastern momentum. Its northern momentum must be 5905 kg*m/s north, since Car 1 had no northern or southern momentum. Using the Pythagorean theorem, we can calculate the overall momentum for Car 2:
p² = (21 114 kg*m/s)² + (5905 kg*m/s)²
p = 21 924 kg*m/s
Let's find the velocity first...we'll worry about the angle later:
p = mv
21 924 kg*m/s = (1500 kg)v
v = 14.6 m/s
Now for the angle. The car's momentum had a northern component of 5905 kg*m/s and an eastern component of 21 114 kg*m/s.
tanθ = (5905 / 21 114)
tanθ = 0.280
θ = arctan(0.280) = 15.6º north of east
I hope that helps. Good luck!
After the collision, the entangled cars (total mass = 2500 kg) were moving at 3.4 m/s at an angle 44º north of east. That's a momentum of:
p = mv = (2500 kg)(3.4 m/s) = 8500 kg*m/s @ 44º north of east
Before we do anything else, let's go ahead and resolve that momentum into its east-west and north-south components:
east-west component: (8500 kg*m/s)cos44º = 6114 kg*m/s east
north-south component: (8500 kg*m/s)sin44º = 5905 kg*m/s north
So...these were the east-west and north-south components of the total momentum BEFORE the collision as well as AFTER the collision.
Now let's examine the cars individually pre-collision. Car 1 (1000 kg) is moving due west at 15 m/s, which gives it a momentum of 15 000 kg*m/s west. It has no north-south momentum. For the sake of simplicity, let's call Car 1's pre-momentum collision -15 000 kg*m/s and the total east-west momentum +6114 kg*m/s.
p1 + p2 = ptotal
(-15,000 kg*m/s) + p2 = +6114 kg*m/s
p2 = 21 114 kg*m/s east
So that's Car 2's eastern momentum. Its northern momentum must be 5905 kg*m/s north, since Car 1 had no northern or southern momentum. Using the Pythagorean theorem, we can calculate the overall momentum for Car 2:
p² = (21 114 kg*m/s)² + (5905 kg*m/s)²
p = 21 924 kg*m/s
Let's find the velocity first...we'll worry about the angle later:
p = mv
21 924 kg*m/s = (1500 kg)v
v = 14.6 m/s
Now for the angle. The car's momentum had a northern component of 5905 kg*m/s and an eastern component of 21 114 kg*m/s.
tanθ = (5905 / 21 114)
tanθ = 0.280
θ = arctan(0.280) = 15.6º north of east
I hope that helps. Good luck!