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geniusboy [140]
3 years ago
5

An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to

the second object.
Physics
1 answer:
garri49 [273]3 years ago
8 0

Answer:

(m1+m2)vo=m1v+m2×0

⟹vo=m1+m2m1v

The impulse imparted to second object is equal to change is momentum is -

J=m2vo=m1+m2m2m1v

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An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

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3 years ago
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Answer:

v = √(25/m)

Explanation:

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v = √(25/m)

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