Question

A rifle with a barrel length of 60 cm fires a 10 g bullet with a horizontal speed of 400 m/s. The bullet strikes a block of wood and penetrates to a depth of 12 cm. . a. what resistive force (assumed to be constant) does the wood exert on the bullet? . b. How long does it take the bullet to come to rest? . c. Draw a velocity verses time graph for the bullet in the wood.

Answer 1

consider the motion of bullet as it penetrates through the block

v₀ = initial velocity of the bullet  = 400 m/s

v = final velocity of the bullet  = 0 m/s

d = distance traveled = 12 cm = 0.12 m

a = acceleration = ?

Using the equation

v² = v₀² + 2 a d

0² = 400² + 2a (0.12)

a = - 6.67 x 10⁵ m/s²

m = mass of the bullet = 10 g = 0.010 kg

The resistive force is given as

f = ma

inserting the values

f = (0.010) (- 6.67 x 10⁵ )

f = - 6670 N

b)

t = time taken by the bullet

using the equation

v = v₀ + a t

0 = 400 + (- 6.67 x 10⁵ ) t

t = 6 x 10⁻⁴ sec

c)

Answer 2

The resistive force of the wood on the bullet is 6,670 N.

The time taken for the bullet to come to rest is 6.0 x 10⁻⁴ s.

The given parameters;

• length of the barrel, L = 60 cm
• mass of the bullet, m = 10 g = 0.01 kg
• speed of the bullet, v = 400 m/s
• depth of penetration, x = 12 cm

The total distance traveled by the bullet in the wood is calculated as follows;

d =  12 cm = 0.12 m

The acceleration of the bullet is calculated as follows;

$v^2 = u^2 + 2as\\\\0 = (400)^2 + 2(0.12) a \\\\-1.44a = 160,000\\\\a = \frac{-160,000}{0.24} \\\\ a = -6.67 \times 10^{5} \ m/s^2$

The resistive force of the wood is calculated as follows;

F = ma

F = 0.01 x (-6.67 x 10⁵)

F = -6,670 N

The time taken for the bullet to come to rest is calculated as follows;

$v= u + at\\\\0 = 400 + (-6.67\times 10^5)t\\\\t = \frac{400}{6.67\times 10^5} \\\\t = 6.0\times 10^{-4} \ s$

The velocity time-graph is presented below.

Learn more here:https://brainly.com/question/19860953