(a) If the KE of an arrow is quadrupled, by what factor has its speed increased?
(b) If the speed is doubled, by what factor does its KE increase?
speed= sqrt(2KE/m)
i don't understand how you are able to find how much it has increased
(a) To find the factor by which the speed of the arrow has increased when the kinetic energy (KE) is quadrupled, we need to understand the relationship between KE and speed.
The kinetic energy of an object is given by the equation KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.
Since we are assuming that the mass of the arrow remains constant, let's denote the initial speed of the arrow as v1 and the increased speed as v2.
According to the given scenario, the initial KE can be represented as KE1 = (1/2)mv1^2, and the increased KE is KE2 = (1/2)mv2^2. We are told that KE2 = 4 * KE1.
Using this information, we can set up the equation: 4 * KE1 = (1/2)mv2^2.
Now, let's cancel out the common factors and simplify the equation: 4 * (1/2)mv1^2 = (1/2)mv2^2.
Next, canceling out the common factors, we get: v1^2 = v2^2.
Taking the square root of both sides, we get: v1 = v2.
Therefore, the speed of the arrow has increased by a factor of 1, which means it remains the same.
(b) To determine the factor by which the kinetic energy (KE) increases when the speed is doubled, we can again use the equation KE = (1/2)mv^2.
Let's assume the initial KE as KE1 and the increased KE as KE2. The initial speed can be denoted as v1, and the doubled speed can be denoted as v2.
Using the given information, we can write KE1 = (1/2)mv1^2 and KE2 = (1/2)mv2^2. We are told that v2 = 2 * v1.
Substituting the values into the equations, we get KE1 = (1/2)m(v1)^2 and KE2 = (1/2)m(2v1)^2 = (1/2)mv1^2(2^2) = (1/2)m(4v1^2) = 2(1/2)mv1^2 = 2KE1.
Therefore, the kinetic energy increases by a factor of 2 when the speed is doubled.