A particle of mass m=9 x 10–31 kg moving towards the wall of a vessel at a velocity of v=600 ms-1 strikes it at an angle of 60° to the normal and rebounds at the same angle at the same speed. The impulse of the force experienced by the wall during the impact is:
Question
A particle of mass m=9 x 10–31 kg moving towards the wall of a vessel at a velocity of v=600 ms-1 strikes it at an angle of 60° to the normal and rebounds at the same angle at the same speed. The impulse of the force experienced by the wall during the impact is:
Solution
The impulse experienced by the wall is equal to the change in momentum of the particle.
First, we need to find the initial and final momentum of the particle.
The initial momentum (p_initial) is given by the mass of the particle times its velocity times the cosine of the angle (because we are interested in the component of the momentum perpendicular to the wall).
p_initial = m * v * cos(60°)
Substituting the given values:
p_initial = 9 x 10^-31 kg * 600 m/s * cos(60°) = 2.7 x 10^-27 kg m/s
The final momentum (p_final) is in the opposite direction but has the same magnitude (because the particle rebounds at the same speed and angle).
p_final = - p_initial = -2.7 x 10^-27 kg m/s
The impulse experienced by the wall is equal to the change in momentum of the particle, which is the final momentum minus the initial momentum:
Impulse = p_final - p_initial = -2.7 x 10^-27 kg m/s - 2.7 x 10^-27 kg m/s = -5.4 x 10^-27 kg m/s
The negative sign indicates that the impulse is in the opposite direction to the initial momentum of the particle.
So, the impulse of the force experienced by the wall during the impact is 5.4 x 10^-27 kg m/s.
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