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3 years ago

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The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a

force of 1.0 x 106N for a period of 20.0 seconds.
a)What is the magnitude of the impulse from the engine?

b)What is the new momentum of the ferry?c)What is the new velocity of the ferry?

2 answers:

goldenfox [79]3 years ago

5 0

Answer:

(a) Impulse of the engine = 20*10^6 N.s

(b) New momentum of the ferry = 1985700 kgm/s

(c) The new velocity of the ferry = 1527.5 m/s

Explanation:

In the given problem, we have:

mass (m) = 13000 kg; velocity (v) = 11 m/s; Force (F) = 1.0*10^6 N; period (t) = 20 s.

From the Newton's law of motion, it is know that:

force*time = mass*velocity; and impulse = force*time

Thus:

(a) the magnitude of the impulse from the engine is:

Impulse = 1.0*10^6 * 20 = 20*10^6 N.s

(b) The new momentum of the ferry is equivalent to the difference between the engine momentum and the ferry momentum. Therefore, we have:

New momentum = Engine momentum - Ferry momentum

Ferry momentum = mass*velocity = 13000*11 =143000 kgm/s

Engine momentum = 1.0*10^6 * 20 = 20*10^6 N.s = 20*10^6 (kgm/s^2 *s) = 20*10^6 kgm/s

Therefore:

New momentum = 20*10^6 - 143000 = 1985700 kgm/s

(c) The new velocity of the ferry is:

v = new momentum/mass = 1985700/13000 = 1527.5 m/s

Pie3 years ago

3 0

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

$J=F.\Delta t$

$J=10^6\ N\times 20\ s$

$J=2\times 10^7\ N-s$

(b) Impulse is also given by the change in momentum i.e.

$J=\Delta p=p_f-p_i$

$J=p_f-p_i$

$p_f=J+p_i$

$p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s$

$p_f=20143000\ kg-m/s$

(c) For new velocity,

$v_f=\dfrac{p_f}{m}$

$v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}$

$v_f=1549.46\ m/s$

Hence, this is the required solution.

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