Question

The figure shows an elastic collision of two pucks (masses \( m_{A}=0.500 \) kg and \( m_{B}=0.300 \mathrm{~kg} \) ) on a frictionless air-hockey table. Puck \( A \) has an initial velocity of \( 4.00 \mathrm{~m} / \mathrm{s} \) in the positive \( x \)-diroction and a find) velocity of \( 2.00 \mathrm{~m} / \mathrm{s} \) in an unknown direction \( a \). Puck \( B \) is intitly at rest. Find the final speed \( v_{B 2} \) of puck \( B \) and the angles \( \alpha \) and \( \beta \). You solved \[ \begin{array}{ll} m_{A}=0.5 \mathrm{~kg} \\ u_{A}=4 \mathrm{im} / \mathrm{s} \\ v_{A}=2 \mathrm{~m} / \mathrm{s} . \end{array} \] (a) conseruation of lainebie erergy. \[ \begin{array}{l} v_{2} m_{A} u_{A}^{2}+y_{2} m_{B} u_{B}^{2}{ }_{2} y_{2} m_{A} v_{A}^{2}+y_{2} m_{B} v_{B}^{2} \\ \therefore v_{B}^{2}=\frac{m_{A} v_{A}^{2}-m_{A} v_{A}^{2}}{m_{B}} \quad\left(=u_{B}=0\right) \\ \therefore v_{B}=\sqrt{\frac{(0.5)(4)^{2}-(0.5)(2)^{2}}{0.3}}=4.47 \mathrm{~m} / \mathrm{s} \end{array} \] (b) By conseruation of momentum lueget. In \( x \)-virection \( m_{A} u_{A 2}=m_{A} V_{A x}+m_{B} V_{B x} \) \[ (0.5)(4)=(0.5)(2)+\cos \alpha)+(0.3)(4.47) \cos \beta \] In \( y \)-Diroction \( 0=m_{A} v_{A y}+m_{B} V_{B y} \quad\left(\because u_{A y}=u_{B y}=0\right) \). \[ 0=(0.5)(2) \sin \alpha+(0.3)(-4.47 \sin \beta) \] By soluing (1) and (2) dueget. \[ \alpha=36.9^{\circ}, \quad \beta=26.6^{\circ} \]

          The figure shows an elastic collision of two pucks (masses \( m_{A}=0.500 \) kg and \( m_{B}=0.300 \mathrm{~kg} \) ) on a frictionless air-hockey table. Puck \( A \) has an initial velocity of \( 4.00 \mathrm{~m} / \mathrm{s} \) in the positive \( x \)-diroction and a find) velocity of \( 2.00 \mathrm{~m} / \mathrm{s} \) in an unknown direction \( a \). Puck \( B \) is intitly at rest. Find the final speed \( v_{B 2} \) of puck \( B \) and the angles \( \alpha \) and \( \beta \).

You solved
\[
\begin{array}{ll}
m_{A}=0.5 \mathrm{~kg} \\
u_{A}=4 \mathrm{im} / \mathrm{s} \\
v_{A}=2 \mathrm{~m} / \mathrm{s} .
\end{array}
\]
(a) conseruation of lainebie erergy.
\[
\begin{array}{l}
v_{2} m_{A} u_{A}^{2}+y_{2} m_{B} u_{B}^{2}{ }_{2} y_{2} m_{A} v_{A}^{2}+y_{2} m_{B} v_{B}^{2} \\
\therefore v_{B}^{2}=\frac{m_{A} v_{A}^{2}-m_{A} v_{A}^{2}}{m_{B}} \quad\left(=u_{B}=0\right) \\
\therefore v_{B}=\sqrt{\frac{(0.5)(4)^{2}-(0.5)(2)^{2}}{0.3}}=4.47 \mathrm{~m} / \mathrm{s}
\end{array}
\]
(b) By conseruation of momentum lueget.

In \( x \)-virection \( m_{A} u_{A 2}=m_{A} V_{A x}+m_{B} V_{B x} \)
\[
(0.5)(4)=(0.5)(2)+\cos \alpha)+(0.3)(4.47) \cos \beta
\]

In \( y \)-Diroction \( 0=m_{A} v_{A y}+m_{B} V_{B y} \quad\left(\because u_{A y}=u_{B y}=0\right) \).
\[
0=(0.5)(2) \sin \alpha+(0.3)(-4.47 \sin \beta)
\]

By soluing (1) and (2) dueget.
\[
\alpha=36.9^{\circ}, \quad \beta=26.6^{\circ}
\]

The figure shows an elastic collision of two pucks (masses mA=0.500 kg and mB=0.300 kg ) on a frictionless air-hockey table. Puck A has an initial velocity of 4.00 m / s in the positive x-diroction and a find) velocity of 2.00 m / s in an unknown direction a. Puck B is intitly at rest. Find the final speed vB 2 of puck B and the angles α and β.

You solved

mA=0.5 kg

uA=4 im / s

vA=2 m / s .

(a) conseruation of lainebie erergy.

v2 mA uA^2+y2 mB uB^22 y2 mA vA^2+y2 mB vB^2
∴ vB^2=(mA vA^2-mA vA^2)/(mB) (=uB=0)
∴ vB=√(((0.5)(4)^2-(0.5)(2)^2)/(0.3))=4.47 m / s

(b) By conseruation of momentum lueget.

In x-virection mA uA 2=mA VA x+mB VB x

(0.5)(4)=(0.5)(2)+cosα)+(0.3)(4.47) cosβ

In y-Diroction 0=mA vA y+mB VB y (∵ uA y=uB y=0).

0=(0.5)(2) sinα+(0.3)(-4.47 sinβ)

By soluing (1) and (2) dueget.

α=36.9^∘, β=26.6^∘

Added by Margaret P.

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