A 4 kg mass is sliding on a fricitonless surface and explodes into two, 2 kg parts. One moving at 3 m/s due north and the other at 5 m/s 30 degrees north of east. What is the original speed of the mass?
Can you please verify if my answer is correct? I used mV_0 = 2(3) =[cos30(5)](2). Then I solved for v_o and got 3.665 m/s. Thanks in advanced for the help.
The original momentum is equal to the final momentum. Determine the final momentum of the parts, and add.
2kg(3N + 5 (cos30 E + sin30 N))
check that. THen, set this to the original momentum 2*velocty, and solve for the vector velocity. I did not punch this into the calculator, but I do not see it equaling what your got
Px = 2(5)cos30 = 8.66
Py = 2(3 + 5sin30) = 11
find magnitude
Ptotal = sqrt(Px²+Py²) = 14
remember p=mv so rearrange the equation.
Vi = Ptotal/Mtotal = 14/4 = 3.5 m/s
To verify if your answer is correct, let's go through the problem step by step.
First, let's determine the final momentum of the two parts. We'll label the mass moving due north as Part A, and the mass moving 30 degrees north of east as Part B.
The final momentum of Part A is given by:
p(A) = mass(A) * velocity(A)
p(A) = 2 kg * 3 m/s (due north)
p(A) = 6 kg·m/s (due north)
The final momentum of Part B is given by:
p(B) = mass(B) * velocity(B)
p(B) = 2 kg * 5 m/s (30 degrees north of east)
Now, we need to break down the velocity of Part B into its north and east components. The north component is given by:
velocity(B)_north = velocity(B) * sin(30 degrees)
velocity(B)_north = 5 m/s * sin(30 degrees)
velocity(B)_north = 5/2 m/s
The east component is given by:
velocity(B)_east = velocity(B) * cos(30 degrees)
velocity(B)_east = 5 m/s * cos(30 degrees)
velocity(B)_east = 5√3/2 m/s
Now, we can calculate the final momentum of Part B:
p(B) = 2 kg (velocity(B)_east + velocity(B)_north)
p(B) = 2 kg (5√3/2 m/s + 5/2 m/s)
p(B) = 2 kg (5√3 + 5)/2 m/s
p(B) = 10 kg (5√3 + 5)/2 m/s
p(B) ≈ 21.6506 kg·m/s (direction is 30 degrees north of east)
Next, we add the final momenta of Part A and Part B to get the original momentum:
p(original) = p(A) + p(B)
p(original) ≈ 6 kg·m/s (due north) + 21.6506 kg·m/s (30 degrees north of east)
To find the magnitude of the original velocity, we could use the Pythagorean theorem to calculate the resultant magnitude, but it seems you've already skipped ahead to assigning it as 2 * v(original). So, let's proceed with that approach.
2 * v(original) ≈ √[(6^2) + (21.6506^2)]
2 * v(original) ≈ √[36 + 468.7669]
2 * v(original) ≈ √504.7669
2 * v(original) ≈ 22.4805
Now, to find v(original), we divide both sides by 2:
v(original) ≈ 22.4805 / 2
v(original) ≈ 11.24025 m/s
Thus, the original speed of the mass is approximately 11.24025 m/s.
It seems your answer, 3.665 m/s, is not correct. Please double-check your calculations.
I hope this explanation helps! Let me know if you have any further questions.