Baseball pitcher throws a fast ball with a 100 Ns impulse. If he applied the force in 0.15 seconds, What force did he apply?

Respuesta :

Answer:

The velocity of the pitcher is 0.105 m/s in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

Therefore,

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

The pitcher has a velocity 0.105 m/s opposite to the direction of the velocity of the ball.

Explanation:

Answer: The force applied on the ball is 666.7 N  

Explanation:

Impulse is defined the change in momentum of an object when force is acted on the object for the given interval of time.

Mathematically,

[tex]\Delta p=F\Delta t[/tex]

where,

[tex]\Delta p[/tex] = change in momentum or impulse = 100 Ns

F = force applied on the ball = ?

[tex]\Delta t[/tex] = time taken = 0.15 s

Putting values in above equation, we get:

[tex]100Ns=F\times 0.15s\\\\F=\frac{100Ns}{0.15s}=666.7N[/tex]

Hence, the force applied on the ball is 666.7 N