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A bullet if mass 20 gram is fired horizontally into a suspended stationary wooden block of mass 380 gram with the velocity of 200 m per second. what is the common velocity of the bullet and the block. If the block and bullet experiences a constant opposing force of 2 Newton , find the time taken to come to rest?

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Final answer:

The student's question involves using conservation of momentum to find the common velocity of the bullet and block after collision, and then using Newton's second law to calculate the time taken for the system to come to rest under a constant opposing force.

Step-by-step explanation:

The student's question addresses a problem involving conservation of momentum and Newton's second law of motion, which are concepts from classical mechanics in physics. To determine the common velocity after the collision, we use the conservation of momentum equation:

Initial total momentum = Final total momentum

(Mass of bullet × Velocity of bullet) + (Mass of block × Velocity of block) = (Total mass) × (Common velocity)

Substituting the given values in the problem, we get:

(0.02 kg × 200 m/s) + (0.38 kg × 0 m/s) = (0.02 kg + 0.38 kg) × Common velocity,

Common velocity = (0.02 kg × 200 m/s) / (0.02 kg + 0.38 kg)

Next, to find the time taken for the block and bullet to come to rest, we apply Newton's second law, which states that force equals mass times acceleration (F = ma). Since we know the opposing force and total mass, we can calculate the deceleration:

Acceleration = Force / Total mass,

Acceleration = -2 N / (0.02 kg + 0.38 kg),

and using the equation of motion, we find the time taken to stop (v = u + at):

0 m/s = Common velocity + (Acceleration × Time),

Solving this equation gives us the time taken for the system to come to rest.

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