Asked by sarah
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
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
Answers
Answered by
Anonymous
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
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
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