A bullet is fired through a wooden board with a thickness of 8.0 cm. The bullet

Question

A bullet is fired through a wooden board with a thickness of 8.0 cm. The bullet hits the board perpendicular to it, and with a velocity of +350 m/s. The bullet then emerges on the other side of the board with a velocity of +210 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration. (Note that the positive velocities define the positive direction to be in the bullet's direction or travel.)

2. Calculate also the total time the bullet is in contact with the board (in sec)

Explanation / Answer

Here ,

a) initial velocity , u = 350 m/s

final velocity , v = 210 m/s

distance , d = 0.08 m

Using third equation of motion

v^2 - u^2 = 2 *a * d

210^2 - 350^2 = 2 * a * 0.08

solving for a

a = -4.9 *10^5 m/s^2

the acceleration of bullet is -4.9 *10^5 m/s^2

b)

using first equation of motion

v = u + a*t

210 = 350 -4.9 *10^5 * t

t = 2.86 *10^-4 s

the total time of contact is 2.86 *10^-4 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.