Answer:
Explanation:
7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s
7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m
or
h = (45sin14.5)² / (2(9.81)) = 6.47 m
which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.
<span>Step 1 -- determine the acceleration of the 200-g block after bullet hits it
a = (coeff of friction) * g
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
a = 0.400*9.8
a = 3.92 m/sec^2
Step 2 -- determine the speed of the block after the bullet hits it
Vf^2 - Vb^2 = 2(a)(s)
where
Vf = final velocity = 0 (since it will stop)
Vb = velocity of block after bullet hits it
a = -3.92 m/sec^2
s = stopping distance = 8 m (given)
Substituting values,
0 - Vb^2 = 2(-3.92)(8)
Vb^2 = 62.72
Vb = 7.92 m/sec.
M1V1 + M2V2 = (M1 + M2)Vb
where
M1 = mass of the bullet = 10 g (given) = 0.010 kg.
V1 = velocity of bullet before impact
M2 = mass of block = 200 g (given) = 0.2 kg.
V2 = initial velocity of block = 0
Vb = 7.92 m/sec
Substituting values,
0.010(V1) + 0.2(0) = (0.010 + 0.2)(7.92)
Solving for V1,
V1 = 166.32 m/sec.
Therefore the answer is (B) 166 m/s!</span>
