Question

An electron in a cathode-ray tube accelerates uniformly from 2.00 3 104 m/s to 6.00 3 106 m/s over 1.50 cm. (a) In what time interval does the electron travel this 1.50 cm? (b) What is its acceleration?

Answer 1

Answer:

(a). The time interval is $4.98\times10^{-9}\ sec$

(b). The acceleration is $1.2\times10^{15}\ m/s^2$

Explanation:

Given that,

Initial velocity $u= 2.003\times10^{4}\ m/s$

Final velocity $v=6.003\times10^{6}\ m/s$

Distance = 1.50 cm

(b) We need to calculate the acceleration

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

s = distance

Put the value into the formula

$(6.003\times10^{6})^2=(2.003\times10^{4})^2+2\times a\times1.50\times10^{-2}$

$a=\dfrac{(6.003\times10^{6})^2-(2.003\times10^{4})^2}{2\times1.50\times10^{-2}}$

$a=1.2\times10^{15}\ m/s^2$

(a). We need to calculate the time interval

Using equation of motion

$v=u+at$

$t=\dfrac{v-u}{a}$

Put the value into the formula

$t=\dfrac{6.003\times10^{6}-2.003\times10^{4}}{1.2\times10^{15}}$

$t=4.98\times10^{-9}\ sec$

Hence, (a). The time interval is $4.98\times10^{-9}\ sec$

(b). The acceleration is $1.2\times10^{15}\ m/s^2$

Answer 2

Answer:

(a) $t=5.02\times 10^{-9}\ s$

(b) $a=1.19\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Initial speed of an electron, $u=2\times 10^4\ m/s$

Final speed of an electron, $v=6\times 10^6\ m/s$

Distance, d = 1.5 cm = 0.015 m

(b) If a is the acceleration of the electron. It is given by the rate of change of its velocity. It can be calculated using third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(6\times 10^6)^2-(2\times 10^4)^2}{2\times 0.015}$

$a=1.19\times 10^{15}\ m/s^2$

(a) Let t is the time interval in which the electron travel 1.50 cm. It is given by :

$t=\dfrac{v-u}{a}$

$t=\dfrac{6\times 10^6-2\times 10^4}{1.19\times 10^{15}}$

$t=5.02\times 10^{-9}\ s$

So, the time interval in which the electron travel 1.50 cm is $5.02\times 10^{-9}\ s$.

Hence, this is the required solution.