First we will find the speed of the ball just before it will hit the floor
so in order to find the speed of the cart we will first use energy conservation
So by solving above equation we will have
now in order to find the momentum we can use
A 500g cart moving at 0.25 m/s collides and sticks to a stationary 750g cart. How fast do the two carts
1 answer:
Answer:
0.1 m/s
Explanation:
Please see attached photo for explanation.
Mass of 1st cart (m₁) = 500 g
Initial velocity of 1st cart (u₁) = 0.25 m/s
Mass of 2nd cart (m₂) = 750 g
Initial velocity of 2nd cart (u₂) = 0 m/s
Velocity (v) after collision =.?
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(500 × 0.25) + (750 × 0) = v(500 + 750)
125 + 0 = v(1250)
125 = 1250v
Divide both side by 1250
v = 125 / 1250
v = 0.1 m/s
Thus, the two cart will move with a velocity of 0.1 m/s after collision.
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a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.