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A 0.145 kg baseball is travelling at 40 $\mathrm{m} / \mathrm{s}$ horizontally when it is struck by a baseball bat. The baseball leaves the bat at 50 $\mathrm{m} / \mathrm{s}$ back in the direction it came from, but at an angle of $40^{\circ}$ above the horizontal. What is the magnitude of the impulse imparted to the baseball?(A) 1.45 $\mathrm{N} \cdot \mathrm{s}$(B) 4.66 $\mathrm{N} \cdot \mathrm{s}$(C) 12.3 $\mathrm{N} \cdot \mathrm{S}$(D) 13.1 $\mathrm{N} \cdot \mathrm{s}$

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Chapter 13

Practice Test 3

Section 1

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Hi in the given problem mask off the baseball. It's given S. M. is equal to 0.14 five kg. The speed with which it hits the bat moving horizontally here this is the direction of motion of this baseball And it is moving with the speed we I and its magnitude is given us 40 meter per second, after which it is reflected back with the speed we have having a magnitude of 50 meter per second, Making an angle of 40° with the horizontal. So it will be having two components one horizontal. Which will be given as we have because 40 degree and you do the opposite direction. It will be taken as negative and another component, vertical component means the component along this bat and that will be we have sign 40 degree. So the initial linear momentum of the baseball will be only along X. Axis means, or we can say along horizontal, but finally linear momentum will be along both directions. X. Axis and Y axis means along horizontal and vertical. Both so we have to find here the impulse imparted by the bed to the ball. So that impulse is actually the change in linear momentum. So to find time pulse magnitude of impulse, we have to find the change in living your momentum means it should be the final linear momentum. B two minus initial linear momentum. Even so, first of all we find the magnitude of changing linear momentum along X axis, then along Y axis. So first of all, delta P. X. Will be given us B. X. F means linear momentum. Final linear momentum along X axis minus initial linear momentum along X axis. So that is given as for final linear momentum, mass of the baseball will be multiplied with this component of velocity along X axis. So it comes out to be minus M. Into we F. Because 40 degrees. And for p excite this is mass of the baseball. But with we I It will be multiplied by the eye. Taking this mask as a comin out along with the negative sign here, this is minus M. We have cost 40 degree minus V. I. So finally plugging in all known values form us. This is 0.145 kilogram for VF. This is 50 in 2000 and 40 degree minus 40 m per second. So finally it comes out to be 0.246 Newton in two seconds. Now, for the change in leader momentum along y axis, it will be final linear momentum along Y axis means P Y F minus initial linear momentum along by excess PV II, which was zero. So for P Y F. It will be given as the product of mass with The vertical component of final velocity means this is we have signed 40°.. So, plugging in unknown values again, this is 0.145 kg for the mask for velocity this is 50 sign 40°.. Need a per second. And finally it comes out to be 4.66 New turn into second. So finally the magnitude of changing linear momentum. Or we can say the impulse that will be given us the square root of Squirtle, B X. Delta P X plus the square of delta P Y. So plugging in The found out values here for Delta P. Ecstasies zero point 246 to the whole square. And for Delta P by this is 4.66 to the whole square. So here it comes out to be 0.06 05 Plus 21.72 Newton in two seconds. Or we can say this is the square root of 21.78 new, turning to second. Or finally this is 4.66 Newton in two seconds. So we can see here our option B is correct. Thank you.

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