A 0.580 kg mass travels along the positive X (horizontal)direction with a veloci

Question

A 0.580 kg mass travels along the positive X (horizontal)direction with a velocity of 17.00 m/s. The particle encounters avertical surface and rebounds with a velocity of 15.0 m/s in thenegative X direction. The time the particle is in contact withthe surface is 3.0 x 10-4 sec. What is the averageforce the surface applies to the particle? A 0.580 kg mass travels along the positive X (horizontal)direction with a velocity of 17.00 m/s. The particle encounters avertical surface and rebounds with a velocity of 15.0 m/s in thenegative X direction. The time the particle is in contact withthe surface is 3.0 x 10-4 sec. What is the averageforce the surface applies to the particle?

Explanation / Answer

   Force   F   =   Rateof change of linearmomentum   =   (final momentum -initial momentum) / time    F   =  ( m* v - m * u) /t      =   m *( v - u) / t    F   =   0.580* { ( - 15.0)   -   17.0} / 3.0 *10-4    F   =   -6.19 * 104   N    Here   - ve sign indicatesthat force acted along -ve X axis.    F   =   -6.19 * 104   N    Here   - ve sign indicatesthat force acted along -ve X axis.

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