a) if a particle's speed decreases by a factor of four, by what factor does its kinetic energy change?
b) particle A has three times the mass and twice the kinetic energy of particle B. What is the speed ratio VA / VB?
a) KE = (1/2) M V^2
What happens to KE when you replace V with a number 4 times larger?
If you can't do that "in your head", try calculating the raio of kinetic energies
KE2/KE1 = (4V)^2/V^2 = ?
b) KEb = KEa/2
KEb/KEa = (Mb/Ma)*(Vb/Va)^2 = 1/2
= (1/3)*(Vb/Va)^2
Solve for Va/Vb
a) To determine how the kinetic energy changes when the speed of a particle decreases, we need to understand the relation between kinetic energy and speed. The formula for kinetic energy (KE) is given by:
KE = (1/2) * m * v^2,
where m represents the mass of the particle and v is the speed.
Now, if the speed decreases by a factor of four, the new speed (v') will be v/4. Plugging this new value into the formula, we can calculate the new kinetic energy (KE'):
KE' = (1/2) * m * (v/4)^2
= (1/2) * m * (v^2 /16)
= (1/16) * m * v^2.
To find the factor by which the kinetic energy changes, we can express the ratio of the new kinetic energy to the original kinetic energy:
KE'/KE = [(1/16) * m * v^2] / [(1/2) * m * v^2]
= [(1/16)] / [(1/2)]
= (1/16) * (2/1)
= 1/8.
Therefore, the kinetic energy changes by a factor of 1/8 when the particle's speed decreases by a factor of four.
b) To determine the speed ratio (VA / VB) between particle A and particle B, we need to consider the relation between kinetic energy, mass, and speed. The formula for kinetic energy (KE) is the same as mentioned above:
KE = (1/2) * m * v^2.
Given that particle A has three times the mass (mA = 3mB) and twice the kinetic energy (KA = 2KB) of particle B, we can equate the two kinetic energy equations:
(1/2) * mA * VA^2 = (1/2) * mB * VB^2.
Since the masses are related through mA = 3mB, we can substitute this into the equation:
(1/2) * (3mB) * VA^2 = (1/2) * mB * VB^2.
Simplifying the equation, we can cancel out the factors of (1/2) and mB, resulting in:
3 * VA^2 = VB^2.
VA^2 / VB^2 = 1/3.
To find the speed ratio (VA / VB), we take the square root of both sides of the equation:
√(VA^2 / VB^2) = √(1/3).
VA / VB = √(1/3).
Hence, the speed ratio between particle A and particle B is √(1/3).