Answer:
a) p = 4.96 10⁻¹⁹ kg m / s
, b) p = 35 .18 10⁻¹⁹ kg m / s
,
c) p_correst / p_approximate = 7.09
Explanation:
a) The moment is defined in classical mechanics as
p = m v
Let's calculate its value
p = 1.67 10⁻²⁷ 0.99 3. 10⁸
p = 4.96 10⁻¹⁹ kg m / s
b) in special relativity the moment is defined as
p = m v / √(1 –v² / c²)
Let's calculate
p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)
p = 4.96 10⁻¹⁹ / 0.141
p = 35 .18 10⁻¹⁹ kg m / s
c) the relationship between the two values is
p_correst / p_approximate = 35.18 / 4.96
p_correst / p_approximate = 7.09
Answer: E = 5.80*10^-13 J
Explanation:
Given
We use the law of conservation of momentum to solve this
Momentum before breakup = momentum after breakup
0 = m1v1 + m2v2
0 = 238m * -2.2*10^5 + 4m * v2
0 = -523.6m m/s + 4m * v2
v2 * 4m = 523.6m m/s
v2 = 523.6 m m/s / 4m
v2 = 130.9*10^5 m/s
v2 = 1.31*10^7 m/s
Using this speed in the energy equation, we have
E = 1/2m1v1² + 1/2m2v2²
E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²
E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]
E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)
E = 9.56*10^-15 + 5.7*10^-13
E = 5.80*10^-13 J
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