Paligutanhk

2021-12-16

A 23.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.00 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 110 g ball swings out to a maximum angle $\theta}_{max}={51.0}^{\circ$

What was$v}_{0$ ?

What was

servidopolisxv

Beginner2021-12-17Added 27 answers

The expression of the initial speed from conservation of energy,

$KE=PE$

$\frac{1}{2}{m}_{b1}{v}_{0}^{2}={m}_{b2}gh$

$\frac{1}{2}{m}_{b1}{v}_{0}^{2}={m}_{b2}gl(1-\mathrm{cos}\theta )$

$v}_{0}=\sqrt{\frac{2{m}_{b2}gl(1-\mathrm{cos}\theta )}{{m}_{b1}}$

Substitute the values,

$v}_{0}=\sqrt{\frac{2\left(110\text{}g\right)\left(9.8\text{}\frac{m}{{s}^{2}}\right)\left(1.00\text{}m\right)[1-\mathrm{cos}\left(51\right)]}{\left(23g\right)}$

$=5.895\text{}\frac{m}{s}$

Answer 5.895 m/s

Substitute the values,

Answer 5.895 m/s

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