Answer:
a ) 4.5 N.s
b) V =5 m/s
Explanation:
given,
mass of rifle(M) = 0.9 kg
mass of bullet(m) = 6 g = 0.006 kg
velocity of the bullet(v) = 750 m/s
a) momentum of bullet = m × v
= 750 × 0.006
= 4.5 N.s
b) recoil velocity
m × u + M × U = m × v + M × V
0 + 0 = 0.006 × 750 - 0.9 × V
V =
V =5 m/s
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>Explanation:</h2>In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
m1= mass 1 = 1.1 kg
Vi1 = initial velocity 1 = 2.7 m/s
m2= 2.4 kg
V2i = -1.9 m/s
We assume east as positive and west as negative.
Apply the formulas:
Vf1 = ?
Replacing:
Answer: 3.6 m/s west