Superman about one half of a second to slow her down to a stop. Assuming that sh

Question

Superman about one half of a second to slow her down to a stop. Assuming that she is traveling at terminal velocity (55 m/s) when Superman catches her and that he applies a constant force to her, I want to know what force Superman has to apply to her during this time. Express your answer as a multiple of her weight. Also, what Impulse does she experience during this time? In an experiment a particle of mass M travels horizontally at V_1=100 m/s and splits into 2 separate pieces m_1 and m_2 as shown below. It is measured experimentally that m_1 = 2m_2, Theta = 20 degree, and Beta 30 Degree. First, find v_1f and v_2f. Second, is Kinetic Energy conserved in this situation? If yes, what kind of collision is it? If no, where do you think that "lost" energy went?

Explanation / Answer

2)
(a)
Vi = 100 m/s
m1 = 2*m2
We know,
m1 + m2 = M
2m2 + m2 = M
m2 = M/3

m1 = 2*m2 = 2* M/3 = 2/3 M

= 20 o
= 30 o

Using Momentum Conservation,
In X axis -

M*vi = m1 * v1f* cos() + m2 * v2f* cos()
M*100 = 2/3*M * v1f* cos(20) + M/3 *v2f* cos(30)
100 = 0.626 v1f + 0.289 * v2f -------1

In Y axis,
m1 * v1f* sin() = m2 * v2f* sin()
2/3*M * v1f* sin(20) = M/3 *v2f* sin(30)
0.684 vif = 0.5 * v2f --------2

Solving eq 1 & 2,

v1f = 97.9 m/s
v2f = 133.94 m/s

(b)
Initial K.E = 1/2*M*100^2 = 5000*M J
Final K.E = 1/2 * m1 * v1f^2 + 1/2*m2*v2f^2
Final K.E = [1/2 * 2/3 * 97.9^2 + 1/2*1/3*133.94^2 ] M
Final K.E = 6184.8*M J

No Energy is not conserved in the collision !!!

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