Question

Use g = 980.35 cm/s^2 foe this problem. a) A 68.44 gram ball is launched with a horizontal speed of 601.78 cm/sec. It is launched from a height of 104.53 cm above the floor. At what distance from the corresponding launch point on the floor docs the ball land? X = b) What is the momentum of the ball when its speed is 601.78 cm/sec? P_ball = (2 numbers after the decimal pt. and in scientific notation) c) Assuming total conservation of momentum, calculate the speed of the ball and pendulum system just after the inelastic collision. The mass of the pendulum is 279.44 grams. Vf v_f = (2 numbers after the decimal pt.) d) After this inelastic collision, what height will the pendulum rise to from its initial position? h = (2 numbers after the decimal pt.) c) Calculate the kinetic energies of the hall and pendulum system before and just after the inelastic collision. K. E_before = K. E_after =

Answer 1

Data Given

g = 980.35 m/s², Horizontal speed u = 601.78 cm/s Vertical height h = 104.53 cm

Part A) Lets first calculate time of flight in case of horizontal projectile

$T = \sqrt{\frac{2h}{g} } =\sqrt{\frac{2\times 104.53}{980.35} }= 0.46 s$

Now

Part B) Initial momentum of the ball will be

$p_{i,ball} = m.u = 68.44g\times 601.78 m/s = 41185.82 g.cm/s$

Part C) Velocity of the ball before collision will be

$v = \sqrt{u^{2}+(gt)^{2}}= \sqrt{(601.78)^{2}+(980.35\times 0.46 )^{2}}$

$v = 752 cm/s$

Now using law of conservation of momentum as

$mv+M.0 = (m+M)v_{f}\Rightarrow v_{f}= \frac{mv}{m+M}$

$v_{f}= \frac{68.44\times 752}{68.44+279.44} = 147.94 cm/s$

Part D) Using conservation of energy after collision

$\frac{1}{2}(m+M)v_{f}^{2}= (m+M)gh$

$h = \frac{v_{f}^{2}}{2g} = \frac{147.94^{2}}{2\times 980.35} = 11.16 cm$

Part E)

KE before collision is

$KE_{before}= \frac{1}{2}mv^{2}= \frac{1}{2}\times 68.44\times 752^{2}=1.935\times 10^{8}erg$

$KE_{after}= \frac{1}{2}(m+M)v_{f}^{2}= \frac{1}{2}\times (68.44+279.44)\times 147.94^{2}=3.81\times 10^{6}erg$