Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
Answer:
Explanation:
Assuming the we have to find ratio maximum forces on the mass in each case
we know that in a spring mass system
F= Kx
K= spring constant
x= spring displacement
Case 1:
case 2:
therefore, 
Answer:
25032.47 W
Explanation:
Power is the time rate of doing work, hence,
P = Work done(non conservative) / time
Work done (non conservative) is given as:
W = total K. E. + total P. E.
Total K. E. = 0.5mv²- 0.5mu²
Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s
Total P. E. = mgh(f) - mgh(i)
Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m
=> W = 0.5mv² - mgh(f)
P = [0.5mv² - mgh(f)] / t
P = [(0.5*790*7²) - (790*9.8*7.2)] / 3
P = (19355 + 55742.4) / 3 = 75097.4/3
P = 25032.47 W
Answer:
Answer is A, it will pass through to focal point after reflecting.
Explanation:
I had the same question in a test, Sorry that you had to do this question in middle school.
