Answer:
Use the equation "KE=½mv²", and use some algebra. > "Particle A has two times the mass...of particle B" mA = 2mB > "Particle A has...8 situations the kinetic power of particle B" KE_A = 8(KE_B) or: ½(mA)(vA)² = 8(½(mB)(vB)²) the rest is uncomplicated algebra: in basic terms sparkling up the above equation for "vA/vB". (hint: start up by utilising dividing the two factors by utilising "(mB)(vB)²". Then make the substitution: mA/mB = 2 (from the 1st eq0.5 ma * VA^2 = 8 * 0.5 * mb * VB^2
0.5 * 0.5*mb * VA^2 = 8 * 0.5 * mb * VB^2
0.5 * 0.5* VA^2 = 8 * 0.5 * VB^2
VA/VB = 4uation))
Explanation:
The ratio of the velocity of particle A to the velocity of particle B is
4 : 1
Let the mass of particle B be m
Let the kinetic energy of particle B be E
Mass of B (m) = m
Kinetic energy of B (KE) = E
[tex]KE = \frac{1}{2}mV_{B}^2\\\\E = \frac{mV_{B}^2}{2} \\\\[/tex]
Cross multiply
[tex]2E = mV_{B}^2[/tex]
Divide both side by m
[tex]V_{B}^2 = \frac{2E}{m} \\\\[/tex]
Take the square root of both sides
[tex]V_{B} = \sqrt{\frac{2E}{m}}[/tex]
Mass of A = ½ mass of B = ½m
Kinetic energy of A (KE) = 8 times KE of B
= 8E
[tex]KE = \frac{1}{2} mV^{2} \\\\8E = \frac{1}{2} * \frac{m}{2} * V_{A}^2 \\\\8E = \frac{mV_{A}^2 }{4}\\\\[/tex]
Cross multiply
[tex]32E = mV_{A}^2[/tex]
Divide both side by m
[tex]V_{A}^2 = \frac{32E}{m}\\\\[/tex]
Take the square root of both sides
[tex]V_{A} = \sqrt{\frac{32E}{m} }\\\\[/tex]
Finally, we shall determine the ratio of velocity of particle A to particle B. This can be obtained as follow:
Velocity of B ([tex]V_{B}[/tex]) = [tex]\sqrt{\frac{2E}{m}}[/tex]
Velocity of A [tex](V_A) = \sqrt{\frac{32E}{m} }\\\\[/tex]
Ratio of [tex]V_A[/tex] To [tex]V_B[/tex] =?
[tex]\frac{V_A}{V_B} = \sqrt{\frac{32E}{m} } : \sqrt{\frac{2E}{m} } \\\\\frac{V_A}{V_B} = \sqrt{\frac{32E}{m} } * \sqrt{\frac{m}{2E} }\\\\\frac{V_A}{V_B} = \sqrt{16}\\\\\frac{V_A}{V_B} = 4 : 1[/tex]
Therefore, the ratio of the velocity of particle A to the velocity of particle B is 4 : 1
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