Question

A proton (mass m 1.67 10 27 kg) is being accelerated along a straight line at 3.6 1015 m/s2 in a machine. If the proton has an initial speed of 2.4 107 m/s and travels 3.5 cm, what then is
(a) its speed and
(b) the increase in its kinetic energy

Answer 1

Answer:

(a) the speed is 2.93 × 10⁷ m/s

(b) the proton's kinetic increased by 2.36 × 10⁻¹³ J

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)

v = 2.93 × 10⁷ m/s

(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)

EK_f - EK_i = 2.36 × 10⁻¹³ J