The final velocity of the bullet+block is 0.799 m/s
Explanation:
We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.
Mathematically, we can write:
where
m = 0.001 kg is the mass of the bullet
u = 800 m/s is the initial velocity of the bullet
M = 1 kg is the mass of the block
U = 0 is the initial velocity of the block (initially at rest)
v is the final combined velocity of the bullet and the block
Solving the equation for v, we find the final velocity:
Learn more about conservation of momentum:
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Explanation:
Consider the kinematic equation,
where x is the distance traveled, v is the initial velocity, a is the acceleration and t is time. By plugging in known values and solving for x,
through simple algebra we get
where this is the distance traveled in meters.
Kinetic energy per unit of mass is
Given,
Therefore,
Now potential energy per unit mass is
Given,
Therefore,
Thus, total mechanical energy of the river water per unit mass is
OR
Answer:
1) The magnitude of the angular acceleration = 67.92 rad/
2) Magnitude of the linear acceleration = 2.744 m/
3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s
4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m
5) the final velocity is 6.21 m/s
Explanation:
the given information :
Bowling mass m = 3.6 kg
Radius = 0.101 m
Initial speed = 8.7 m/s
Coefficient of kinetic friction μ = 0.28
1) he magnitude of the angular acceleration of the bowling ball is
F = m a
= μ N , N = m g
= μ m g
1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:
momen inersia of Bowling ball I = (2/5) m
torque τ = I α
τ = F R
I α = F R
(2/5) m α = μ m g R
α = (5 μ g / 2R) μ g R
= (5 x 0.28 x 9.8/ 2 x 0.101)
= 67.92 rad/
2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane
F = - ,
is the force of kinetic friction
m a = - μ m g, remove m
the magnitude of linear accelaration is
a = μ g
= (0.28) (9.8)
= 2.744 m/
3) The bowling ball takes time to begin rolling without slipping:
The linear speed, =
- a t
=
- μ g t
the angular speed, ω = ω0 + α t
ω = ω0 + (5 μ g/2R ) t
= ω R
- μ g t = ω0 R + (5 μ g/2R ) t R
7 μ g t/2 = + ω0 R
hence,
t = (2 + ω0 R)/ 7 μ g
ω0 = 0 (no initial spin), therefore
t = 2 / 7 μ g
= 2 x 8.7 / 7 (0.28) (9.8)
= 0.906 s
4) How long it takes for the bowling ball to begin rolling without slipping, S
S = t - (1/2) a
= (8.7) (0.906) - (1/2) (2.744)
= 6.75 m
5) The final velocity
=
- a t
= 8.7 - (2.744) (0.906)
= 6.21 m/s