A 100gm ball moving horizontally with 20m/s is struck by a bat, as a result it starts moving with a speed of 35m/s at an angle of 37 above the horizontal in the same verticle plane as shown in the figure. ltb/rgt (a) Find the average force exerted by the bat if duration of impact is 0.30s.ltb/rgt (b) FInd the average force exerted by the bat if duration of impact is 0.03s. (c) Find the average force exerted by the bat if duration of impact is 0.003s. (d) What do you conclude for impulse of weight of the ball as duration of contact decreases?

Explanation

The impulse momentum diagram of the ball is dhown in the figure below. Here F, mg, and Deltat represent the average value of the force exerted by the bat, weight of the ball and time interval.

Applying principle of impulse and momentum in x-direction, we have ltbr gt `p_(fx)=p_(ix)+sum l_(mp,x) rarr " " -2+F_xDeltat=2.8`
`F_x=4.8/(Delta t)N`...(i)
Applying principle of impulse and momentum in y-direction, we have
`p_(fy) = p_(iy)+sum I_(mp.y) rarr " " 0.0 + F_(y)Deltat - 1.0Delta t = 2.1`
`F_(y) = ((2.1)/(Deltat)+1.0)N`.....(ii)
Applying principle of impulse and momentum in y-direction, we have
(a) Substituting `Delta t = 0.30s`, on equations (i) and (ii), we find `vec(F) = 16hat(i) + 8hat(j) N`
(b) Substituting `Deltat = 0.03s`, in equations (i) and (ii), we find `vec(F) = 160hat(i) + 71hat(j) N`
Substituting `Deltat = 0.003s`, in equations 1 and 2, we find `vec(F) = 1600hat(i) + 701hat(j) N`
(d) It is clear from the above results that as the duration of contact between the ball and the bat decreases, effect of the weight of the ball also decreases as compared with that of the force of the bat and for sufficiently short time interval, it can be neglected.