Question

mass of bullet is 0.03kg fire 200 in a sec fire with 50 m/s and reflect back with 30m/s​

Answer 1

Mass of the body = 0.03 kg .

Final momentum = 200 × 0.03 kg × (-30 m/s)

⇒ - 2 × 3 × 30 kg m/s

⇒ - 6 × 30 kg m/s

⇒ - 180 kg m/s

Initial momentum = 200 × 0.03 kg × 50 m/s

⇒ 2 × 3 × 50 kg m/s

⇒ 6 × 50 kg m/s

⇒ 300 kg m/s

Change in momentum = 300 kg m/s - ( -180 kg m/s )

⇒ 480 kg m/s .

Average force acting = change in momentum / time taken

⇒ 480 kg m/s / 1 s

⇒ 480 kg m/s²

⇒ 480 N

Answer 2

Answer:

[Mass of the bullet = 0.03 kg .]

Final momentum = 200 × 0.03 kg × (-30 m/s)

⇒ - 2 × 3 × 30 kg m/s

⇒ - 6 × 30 kg m/s

⇒ - 180 kg m/s

Initial momentum = 200 × 0.03 kg × 50 m/s

⇒ 2 × 3 × 50 kg m/s

⇒ 6 × 50 kg m/s

⇒ 300 kg m/s

Change in momentum = 300 kg m/s - ( -180 kg m/s )

⇒ 480 kg m/s .

Average force acting = change in momentum / time taken

⇒ 480 kg m/s / 1 s

⇒ 480 kg m/s²

⇒ 480 N

Explanation:

Momentum = mass × velocity .