Magnitude of the baseball's momentum?

[tnutty]

Homework Statement

A cardinal (Richmondena cardinalis) of mass 3.80×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is the ratio of the cardinal's magnitude of momentum to the magnitude of the baseball's momentum?

p_c / p_b = ?


attempt :

P_c / p_b = M_c*V_c / M_b*V_b

M_c / M_b = 19/72 ; their ratio

so their V has to be a ratio of = V_c/V_b = 72/19.

not sure what after this.

 

[rock.freak667]

$$\frac{0.5m_cv_c^2}{0.5m_bv_b^2}=1$$
{since they have the same kinetic energy)
(b=basketball)

you can get m_c/m_b, so can get the ratio of the velocites.

using the momentum equation you can get the ratios of the momentums

$$\frac{p_c}{p_b}=\frac{m_cv_c}{m_bv_b}$$

 

[tnutty]

Confused a little.

$$\frac{0.5m_cv_c^2}{0.5m_bv_b^2}=1$$

the .5 cancels out. I know M_c / M_b

and I use
$$\frac{p_c}{p_b}=\frac{m_cv_c}{m_bv_b}$$

SO Do i find the ratio of velocities and plug it into momentum ratio?

 

[tnutty]

[attempt]

If I find the ratio of the velocity with $$\frac{0.5m_cv_c^2}{0.5m_bv_b^2}=1$$
then put the velocity ratio into :
$$\frac{p_c}{p_b}=\frac{m_cv_c}{m_bv_b}$$

the ratio of momentum = 1.

so i am not getting what your saying

 

[rl.bhat]

Ec = 1/2*mc*vc^2
2*Ec*mc = (mc^2)*vc^2 = Pc^2...(1)
Similarly 2*Eb*mb = (mb^2)*vb^2 = Pb^2...(2)
Now take the ratio of equation 1 and 2 and find the ratio of the momentums.

 

[sArGe99]

$$K.E = \frac{p^2}{2m}$$

$$\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}<br />$$

$$\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}<br />$$