A 3.40 kg steel ball strikes a wall with a speed of 12.0 m/s at an angle of 60.0° with the surface. It bounces?
A 3.40 kg steel ball strikes a wall with a speed of 12.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball? (Assume right is the positive direction.)
=____i + ____J
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Hi,
The average force (Fm), by means of data of the problem may be obtained by variation linear momentum (Dp) in the time interval (Dt) given.
Fm = Dp / Dt, but
only the momentum component perpendicular to the wall will
help (transfer momentum to the wall), so
Dpx = m * V2 * cos (theta) - (- m * V1 * cos (theta)),
perpendicular component
Dpy = m * V2 * sin(theta) - (m * V1 * sin(theta)) = 0, parallel component
the negative sign in the second term (Dpx) is because the
orientation of the trajectory.
^ (y direction)
|
| -------------------> M * V2 * cos (theta)
| ------------------------------------------------------------> positive direction (x direction)
| <------------------M * V1 * cos (theta)
Substituting the values we have: (note that V2=V1=12.0m/s):
Fm = 2*m*V*cos(theta)/Dt = 2*3.20*12.0*cos( 60.0°)/.200 = 3600 N
Fm = 3.6 kN
Vectorial form:
Fm = 3.6kN i + 0 j
Hope I help