Question #e3483

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Answer 1 · 2015-07-19T16:18:23

Momentum will decrease by 20%.

The idea behind this problem is that kinetic energy and momentum both use the velocity of the object in their expressions.

The equation for an object's kinetic energy looks like this

$E_K = 1/2 * m* v^2$ , where

$m$ - the mass of the object;
$v$ - its velocity.

The equation that describes the momentum of an object is

$p = m * v$

So, you know that the object's kinetic speed decreases by 36%. This means that it now stands at 64% of its initial value.

Since the mass of the object is assumed to be constant, this decrease in kinetic energy correlates with a decrease in velocity.

If $v_i$ is the initial velocity and $v_f$ is the velocity after the kinetic energy is reduced, then you can write

$E_("K f") = 64/100 * E_("K i")$

$cancel(1/2 * m) * v_f^2 = cancel(1/2 * m) * 0.64 * v_i^2$

Take the square root from both sides of the equation to get

$sqrt(v_f^2) = sqrt(0.64 * v_i^2)$

$v_f = 0.8 * v_i$

Now use this relationship between the two velocities of the object to express the initial and final momentum

$p_i = m * v_i$

$p_f = m* v_f = m * 0.8 * v_i = 0.8 * underbrace(m * v_i)_(color(blue)(=p_i)) = 0.8 * p_i$

This means that the momentum decreased by

$Delta_p = p_i - p_f = p_i * (1 - 0.8) = 0.2 * p_i$

Expressed in percentages, this is equivalent to

$Delta_p/p_i = (0.2 *cancel(p_i))/cancel(p_i) * 100 = color(green)(20%)$