<p>Can someone confirm my answer for a physics-related problem?</p>
<p>Here it is:</p>
<p>mass of golf ball = 0.046 kg
mass of club = 0.22 kg</p>
<p>A golf ball resting on a tee is struck by club with a speed of 44 m/s.
Find the speed of the ball when it leaves the tee.</p>
<p>I got 210 m/s. Someone please confirm!</p>
<p>I personally think that this is a bad problem, or information was not included, but assuming that the club stops at the very instant it hits the ball and, therefore, transfers all of its momentum</p>
<p>you would simply do </p>
<p>mv = mv</p>
<p>so 44 * 0.22 = v * 0.046</p>
<p>v = 210</p>
<p>However, if this was a "good" problem, you would be given the length of the club, the speed at which the club is moving at the point at which the golfer is holding it, the position at which the golfer is holding the club (therefore forcing you to do angular momentum/velocity and cetripetal forces), the impulse of the shot (you could ask for the change in momentum and subsequent velocity of the ball). Then to make it more fun, you could include the angle of the club face, the position of the ball which was struck and have to solve the trajectory of the ball and the distance traveled. While on that trajectory, you could add a cross wind and position of the hole, given the golfer gets a hole in one, how fast was the wind blowing ....</p>
<p>wow...thats all i have to say</p>
<p>thanks. yeah i thought this was a trick question or something.</p>
<p>if its elastic then u dont need the final v for the club, but if its not, then it has to be determinable</p>
<p>okay another problem that i need someone to confirm:</p>
<p>A 0.000150 kg bullet hits a 1.00 kg block horizontally at a velocity of 340 m/s. All of the bullet's momentum is transferred over to the block. The block is resting on a ramp (think Superman at Six Flags). How high does it go? (Not how far). Oh yeah, no friction.</p>
<p>I got 1.33 x 10^-4 m</p>
<p>Let's start off with Conservation of Momentum.
Pi=Pf
mv +mv=(M+m)v
(0.000150)(340)+0=(0.000150+1)(v)
V after collision(block and bullet system0=0.051m/s</p>
<p>If we label the distance the block system moves up the ramp X, then we could relate how high the system goes with how far it moves up the ramp, resulting in X=h/sin(theta) after we consider basic trig concepts.</p>
<p>Next, we must uses forces to find what the acceleration of the block-bullet system is. We only need to worry about the forces in the x-direction i nthsi case...
mgsin(theta)=ma
a=gsin(theta)</p>
<p>Finally, we can now use a Kinematic Equation to find h...
v^2=Vo^2+2a[h/sin(theta)]
0=(0.051)^2+2[gsin(theta)][h/sin(theta)]
0=(0.051)^2+2(9.8)h
h=1.33*10^-4m</p>
<p>Yep, you apparently solved the question correctly since I got the same answer haha. Have more self-confidence!!!:)</p>
<p>Peace.</p>
<p>Thanks evil. Yeah, I really need some of that!</p>