Question

a particle strikes a sphere at angle 37 with the radius particle mass 1kg velocity 10 ms and sphere mass 5kg rest what is the force exerted on the sphere in 10 seconds

Answer 1

4/3 N

Answer 2

To solve this problem, we need to understand the concept of impulse and how it relates to average force. The question asks for the "force exerted on the sphere in 10 seconds," which is an unusual phrasing for a collision. The most plausible interpretation is that it asks for the average force over a 10-second period, where the total impulse delivered to the sphere during the collision is spread out over this time.

Here's the step-by-step solution:

1. Resolve the initial velocity of the particle:
   The particle strikes the sphere at an angle of $37^\circ$ with the radius. The radius represents the normal to the surface at the point of impact.
   Initial velocity of particle, $v_1 = 10$ m/s.
   Angle with normal, $\theta = 37^\circ$.
   Normal component of particle's initial velocity: $v_{1n} = v_1 \cos \theta = 10 \cos 37^\circ$.
   Since $\cos 37^\circ \approx 0.8$, $v_{1n} = 10 \times 0.8 = 8$ m/s.
   Tangential component of particle's initial velocity: $v_{1t} = v_1 \sin \theta = 10 \sin 37^\circ$.
   Since $\sin 37^\circ \approx 0.6$, $v_{1t} = 10 \times 0.6 = 6$ m/s.

2. Apply collision principles:
   We assume the collision is elastic ($e=1$) as no coefficient of restitution is given, and the surface is smooth (no friction), so the tangential component of the particle's velocity remains unchanged.
   Initial velocity of sphere, $v_2 = 0$ m/s. Its normal component $v_{2n} = 0$.

   • Conservation of momentum along the normal direction:
     $m_1 v_{1n} + m_2 v_{2n} = m_1 v_{1n}' + m_2 v_{2n}'$
     Given $m_1 = 1$ kg, $m_2 = 5$ kg.
     $1(8) + 5(0) = 1(v_{1n}') + 5(v_{2n}')$
     $8 = v_{1n}' + 5v_{2n}'$ (Equation 1)

   • Coefficient of restitution ($e=1$) along the normal direction:
     $e = \frac{v_{2n}' - v_{1n}'}{v_{1n} - v_{2n}}$
     $1 = \frac{v_{2n}' - v_{1n}'}{8 - 0}$
     $8 = v_{2n}' - v_{1n}'$ (Equation 2)

3. Solve for final velocities:
   From Equation 2, $v_{1n}' = v_{2n}' - 8$.
   Substitute this into Equation 1:
   $8 = (v_{2n}' - 8) + 5v_{2n}'$
   $8 = 6v_{2n}' - 8$
   $16 = 6v_{2n}'$
   $v_{2n}' = \frac{16}{6} = \frac{8}{3}$ m/s.

   This is the final normal velocity of the sphere. The sphere gains velocity only in the normal direction.

4. Calculate the impulse on the sphere:
   Impulse ($J$) is the change in momentum of the sphere.
   $J = \Delta p_{sphere} = m_2 v_{2n}' - m_2 v_{2n}$
   $J = 5 \times \frac{8}{3} - 5 \times 0$
   $J = \frac{40}{3}$ Ns.

5. Calculate the average force:
   The question asks for the force exerted on the sphere in 10 seconds. This is interpreted as the average force over a 10-second interval, given the total impulse.
   Average Force ($F_{avg}$) = $\frac{\text{Impulse}}{\text{Time Interval}}$
   $F_{avg} = \frac{J}{\Delta t}$
   $F_{avg} = \frac{40/3}{10}$
   $F_{avg} = \frac{40}{30} = \frac{4}{3}$ N.

The force exerted on the sphere in 10 seconds, interpreted as the average force, is $\frac{4}{3}$ N.