Answer:
(a) the speed is <em>2.93 × 10⁷ m/s </em>
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(b) the proton's kinetic increased by <em>2.36 × 10⁻¹³ J</em>
Explanation:
The given information is:
- initial speed, v_i = 2.4×10⁷ m/s
- distance travelled, d = 0.035 m
- acceleration, a = 3.6×10¹⁵ m/s²
- mass, m = 1.67×10⁻²⁷ kg
(a) We must first determine the time it took the proton to travel the given distance:
t = d / v_i
t = (0.035 m) / (2.4×10⁷ m/s)
t = 1.46 × 10⁻⁹ s
Therefore, using the equation kinematics, we can determine the speed of the proton after 1.46 × 10⁻⁹ seconds. The speed is:
v = v_i + a t
v = (2.4 × 10⁷ m/s) + (3.6×10¹⁵ m/s²)(1.46 × 10⁻⁹ s)
<em>v = 2.93 × 10⁷ m/s </em>
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(b) We must determine the inertial kinetic energy and the final kinetic energy:
The initial kinetic energy is:
EK_i = 1/2 m v_i²
= 1/2(1.67 × 10⁻²⁷ kg)(2.4 × 10⁷ m/s)²
= 4.81 × 10⁻¹³ J
The final kinetic energy is:
EK_f = 1/2 m v_f²
= 1/2(1.67 × 10⁻²⁷ kg)(2.93 × 10⁷ m/s)²
= 7.17 × 10⁻¹³ J
Therefore, the proton's kinetic increased by:
EK_f - EK_i = (7.17 × 10⁻¹³ J) - (4.81 × 10⁻¹³ J)
<em>EK_f - EK_i = 2.36 × 10⁻¹³ J</em>
