Question

The nucleus of a helium atom travels along the inside of a straight hollow tube 4 m long which forms part of a particle accelerator. If one assumes uniform acceleration, how long is the particle in the tube if it enters at a speed of 2000 m/s and leaves at 8000 m/s?

• A. \( 0.6 \times 10^{-4} \, \text{s} \)
• B. \( 4 \times 10^{-4} \, \text{s} \)
• C. \( 8 \times 10^{-4} \, \text{s} \)
• D. \( 3.3 \times 10^{-4} \, \text{s} \)

Hint:

To solve for time or acceleration under uniform acceleration, use the kinematic equations: \( v^2 = u^2 + 2as \) and \( v = u + at \).

Answer 1

The Correct Option is C

We need to use the equations of motion under uniform acceleration to solve this. The kinematic equation we will use is: \[ v^2 = u^2 + 2as \] where: - \( v \) is the final velocity (8000 m/s), - \( u \) is the initial velocity (2000 m/s), - \( a \) is the acceleration, - \( s \) is the distance traveled (4 m). Rearranging to solve for acceleration \( a \): \[ a = \frac{v^2 - u^2}{2s} \] Substituting the known values: - \( v = 8000 \, \text{m/s} \), - \( u = 2000 \, \text{m/s} \), - \( s = 4 \, \text{m} \), \[ a = \frac{(8000)^2 - (2000)^2}{2 \times 4} \] \[ a = \frac{64000000 - 4000000}{8} = \frac{60000000}{8} = 7500000 \, \text{m/s}^2 \] Now, using the equation \( v = u + at \) to solve for time \( t \): \[ t = \frac{v - u}{a} \] Substituting the known values: \[ t = \frac{8000 - 2000}{7500000} \] \[ t = \frac{6000}{7500000} = 8 \times 10^{-4} \, \text{s} \] Thus, the time the particle spends in the tube is \( 8 \times 10^{-4} \, \text{s} \).