The figure shows a 0.3 kg baseball just before and just after it collides with a

Question

The figure shows a 0.3 kg baseball just before and just after it collides with a bat. Just before, the ball has velocity v 1 of magnitude 11.5 m/s and angle ?? = 38.0°. Just after, it is traveling directly upward with velocity v 2 of magnitude 9.00 m/s. The duration of the collision is 1.00 ms. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat? (a) Numbe (b) Numbe (c) Numbe (d) Numbe nits [kg.m/s or N-s unies IJni unice Uiii,

Explanation / Answer

Ans:-

Given data :-mb = 0.3kg , v1= 11.5m/s, ?1 = 38deg

V2 = 9m/s, ?2 = 90deg , ?t= 1ms

Impulse J = ?P

So calculate initial and final momentum

x-axis

Pi =m*Vx = 0.3*-11.5*cos38=-2.72kgm/s

Pf = 0kgm/s

Y-axis

Pi = 0.3 * -11.5*sin38 =-2.12kgm/s

Pf = 0.3*12 =3.6kgm/s

?Px = Pfx-Pix =2.72kgm/s

?Py = 5.72kgm/s

J= ?P = sqrt(2.72^2 + 5.72^2) = 6.33kgm/s

? = tan^-1 (?Py/?Px) = tan^-1 (5.72/2.72) = 64.57deg

J = F*?t

J/?t = F

F= 6.33/ 1*10^-3 = 6.33*10^3N

Direction is same 64.57deg from xaxis

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---