A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo
1 answer:
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
In case of perfectly inelastic collision v'1 and v'2 are same.
We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁
Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg
Mass of second mud ball is 6kg.
In case of perfectly inelastic collision v'1 and v'2 are same.
We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁
Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg
Mass of second mud ball is 6kg.
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a) Time of flight: 22.6 s
To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.
The vertical position at time t is given by
where
h = 2.5 km = 2500 m is the initial height
is the initial vertical velocity of the cargo
g = 9.8 m/s^2 is the acceleration of gravity
The cargo reaches the ground when
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
b) 7.5 km
The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:
So the horizontal distance travelled is
And if we substitute the time of flight,
t = 22.6 s
We find the range of the cargo:
a) The impulse is 76.5 Ns
b) The average force is 546.4 N
c) The final speed is 31.5 m/s
Explanation:
a)
The impulse exerted on an object is defined as
where
F is the magnitude of the force exerted on the object
is the time interval during which the force is applied
If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.
Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are
and whose height is
Therefore, the area (and the impulse) is
b)
In this problem, the force applied is not constant. However, we can rewrite the impulse also as
where
is the average force exerted during the whole time
In this problem we have
J = 76.5 Ns is the impulse (calculated in part a)
is the time interval
Solving for the average force, we find
c)
According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:
where
m is the mass of the object
v is the final velocity
u is the initial velocity
In this problem, we have
J = 76.5 Ns
m = 3.0 kg is the mass
u = 6.0 m/s is the initial velocity
Solving for v, we find the final velocity (and speed):
Learn more about impulse and momentum:
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