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A bullet of mass 10 g travelling horizontally with a velocity ... - Science
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s⁻¹ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
The calculations:
Distance of penetration:
Given:
Mass (m) = 10 g = 0.01 kg
Initial velocity (u) = 150 m/s
Final velocity (v) = 0 m/s (comes to rest)
Time (t) = 0.03 s
First, calculate the acceleration (a) using the formula v = u + at:
0 = 150 + a * 0.03
-150 = 0.03a
a = -150 / 0.03 = -5000 m/s²
Now, calculate the distance (s) using the formula s = ut + ½at²:
s = (150 * 0.03) + (½ * -5000 * (0.03)²)
s = 4.5 + ( -2500 * 0.0009)
s = 4.5 - 2.25
s = 2.25 m
Alternatively, using v² = u² + 2as:
0² = 150² + 2 * (-5000) * s
0 = 22500 - 10000s
10000s = 22500
s = 22500 / 10000 = 2.25 m
The distance of penetration of the bullet into the block is 2.25 m.
Magnitude of the force:
Using Newton's second law, F = ma:
F = 0.01 kg * (-5000 m/s²)
F = -50 N
The magnitude of the force exerted by the wooden block on the bullet is 50 N. (The negative sign indicates the force is in the opposite direction of the bullet's motion).