Question

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 6.10 × 1010 m/s2 in a machine. If the proton has an initial speed of 7.50 × 105 m/s and travels 1.50 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

Answer 1

a)

$$v^2=v_0^2+2as$$

$$v^2=(7.5\cdot10^5)^2+2(6.1\cdot10^{10})(0.015)$$

$$v=7.51\cdot10^5\frac{m}{s}$$

b)

$$\Delta K=0.5m(v^2-v_0^2)=0.5m(2as)=mas$$

$$\Delta K=(1.67\cdot10^{-27})(6.1\cdot10^{10})(0.015)=1.53\cdot10^{-18}\ J$$