Answer to Question #25160 in Mechanics | Relativity for Patrick
v_1 = sqrt(2*g*h_1);
In the similar fashion, just after rebounding from the floor, v_2 = sqrt(2*g*h_2), where h_2 is the height reached after rebounding.
If we define the direction away from the floor (up) as positive and towards the floor (down) as negative, and project the velocities on this axis,
v_1 = - sqrt(2*g*h_1);
v_2 = sqrt(2*g*h_2).
The change of momentum, taking account of the change of direction, is expressed as
dp = p2 - p1 = m*v_2 - m*v_1 = m * ( sqrt(2*g*h_1) + sqrt(2*g*h_2) ).
Substituting values, we get
dp = 0.150 kg * ( sqrt(2*(9.806 m/s^2)*(1.25 m)) + sqrt(2*(9.806 m/s^2)*(0.96
m)) ) = 1.3935 kg*m/s
Answer: impulse of about 1.4 kg*m/s
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