The block with the bullet lodged in the block is now travelling at 2.133 m/s.
<h3>What is momentum conservation principle?</h3>
When there is no external force acting on the system, the momentum remains conserved.
For inelastic collision, after collision both objects travel with common speed.
m1u1 + m2u2 =(m1 +m2)v
Substitute initial velocity of bullet u1 =320 m/s , initial velocity of block u2 =0, mass of bullet m1 = 0.1 kg and mass of block m2 = 14.9 kg.
Solve for the final velocity of bullet,
0.1 x 320 + 14.9 x 0 = (0.1 +14.9) x v
v = 2.133 m/s
Thus, the block with the bullet lodged in block now travelling at 2.133 m/s.
Learn more about momentum conservation principle.
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Explanation:
It is given that,
The acceleration of a particle, 
(negative as the particle is decelerating)
Initial distance, x₁ = 20 m
Initial time, t₁ = 4 s
New distance x₂ = 4 m
Velocity, v = 10 m/s
(A) Calculating initial distance using second equation of motion as :
u = 21 m/s
When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :
t = 2.62 seconds
So, the velocity of the particle is zero at t = 2.62 seconds.
(B) Velocity at t = 11 s
v = 13 m/s
Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.
d = 132.48 m
So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.
Answer:
it means how fast the wave is traveling in which a 40 ms is as fast as a car traveling
