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Naily [24]
3 years ago
5

A constant net force is applied to an object. Suddenly, the object's mass doubles, but the net force remains the same. For this

to happenthe acceleration must do what?
A:Quadruple
B:Double
C:Stay the same
D:Halve
Physics
1 answer:
____ [38]3 years ago
8 0

<u>D: Half</u>

We know that,

F = m.a, where F is the force, m is the object's mass and a is the acceleration.

In the first case, we observed that a1 = F/m.

In the second case, we observed that the mass has been doubled, so a2 = F/2m .

By the ratio of the two cases, we get

a1/a2 = F/m / F/2m

or, a1/a2 = 2

or, a1 = 2.a2

or, a1/2 = a2

Therefore, the acceleration gets <u>half</u> of it's original measurement.

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a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially
CaHeK987 [17]

Answer:

Velocity of the helium nuleus  = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

P_{1i} + P_{2i} = P_{1f} + P_{2f] (1)

<em>where P_{1i}: is the proton linear momentum initial, P_{2i}: is the helium nucleus linear momentum initial, P_{1f}: is the proton linear momentum final, P_{2f}: is the helium nucleus linear momentum final </em>

<u>From (1):</u>

m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, v_{1i} and v_{2i}: are the proton and helium nucleus velocities, respectively, before the collision, and v_{1f} and v_{2f}: are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

K_{1i} + K_{2i} = K_{1f} + K_{2f} (3)

<em>where K_{1i} and  K_{2i}: are the kinetic energy for the proton and helium, respectively, before the colission, and K_{1f} and  K_{2f}: are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} (4)  

<u>Now we have two equations: (2) ad (4), and two incognits: v_{1f} and v_{2f}. </u>

Solving equation (2) for v_{1f}, we have:

v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} (5)

<u>From getting (5) into (4) we can obtain the v_{2f}:</u>

v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0

v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}

v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

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