Respuesta :
Answer:
The resultant speed of 50 gm blob of clay is [tex]V_{resulatnt} = 0.4[/tex] [tex]\frac{m}{s}[/tex]
and Direction is [tex]\theta[/tex] = 63.86°
Explanation:
Given data
[tex]m_1 = 20 gm[/tex]
[tex]V_{1x}[/tex] = 2 [tex]\frac{m}{s}[/tex]
[tex]V_{1y}[/tex] = 0
[tex]m_2[/tex] = 30 gm
[tex]V_2[/tex] = 1.2 [tex]\frac{m}{s}[/tex]
[tex]\theta =[/tex] 30
[tex]V_{2x}[/tex] = - 1.2 [tex]\cos 30[/tex] = - 1.039 [tex]\frac{m}{s}[/tex]
[tex]V_{2y} = - 1.2 \sin 30[/tex] = - 0.6 [tex]\frac{m}{s}[/tex]
From conservation of momentum principal in x- direction
[tex]V_{1x}[/tex] [tex]m_{1}[/tex]+ [tex]m_2[/tex] [tex]V_{2x}[/tex] = ( [tex]m_1[/tex] + [tex]m_2[/tex] ) [tex]V_{fx}[/tex]
Put all the values in above equation we get
20 × 2 + 30 × ( -1.039 ) = (30 + 20) [tex]V_{fx}[/tex]
[tex]V_{fx}[/tex] = 0.1766 [tex]\frac{m}{s}[/tex]
From conservation of momentum principal in y - direction
[tex]V_{1y}[/tex] [tex]m_{1}[/tex]+ [tex]m_2[/tex] [tex]V_{2y}[/tex] = ( [tex]m_1[/tex] + [tex]m_2[/tex] ) [tex]V_{fy}[/tex]
20 × 0 + 30 × ( - 0.6 ) = (30 + 20) [tex]V_{fx}[/tex]
[tex]V_{fy}[/tex] = - 0.36 [tex]\frac{m}{s}[/tex]
Now resultant velocity
[tex]V_{resulatnt} = \sqrt{V_{fx} ^{2} +V_{fy} ^{2} }[/tex]
[tex]V_{resulatnt} = \sqrt{(0.1766) ^{2} +(-0.36) ^{2} }[/tex]
[tex]V_{resulatnt} = 0.4[/tex] [tex]\frac{m}{s}[/tex]
Direction
[tex]\theta = tan^{-1} (\frac{V_{fy} }{V_{fx} })[/tex]
[tex]\theta =tan^{-1} \frac{-0.36}{0.1766}[/tex]
[tex]\theta[/tex] = 63.86°
Therefore the resultant speed of 50 gm blob of clay is [tex]V_{resulatnt} = 0.4[/tex] [tex]\frac{m}{s}[/tex]
and Direction is [tex]\theta[/tex] = 63.86°
The speed of the resulting 5 g balls is 1.076 m/s and the direction is 42⁰.
The given parameters;
- mass of the ball, m = 20 g = 0.02 kg
- initial velocity of the ball, u = 2 m/s
- mass of the second ball, m = 30 g = 0.03 kg
- initial velocity of the second ball, u = 1.2 m/s
The resultant initial momentum of the two balls is calculated as follows;
[tex]P_i = \sqrt{P_i_x^2 + P_i_y^2} \\\\P_i = \sqrt{(0.02 \times 2)^2 + (0.03\times 1.2)^2} \\\\P_i = 0.0538 \ kgm/s[/tex]
Apply the principle of conservation of linear momentum to determine the final velocity of the balls;
[tex]P_f = P_i\\\\v(0.05) = 0.0538\\\\v = \frac{0.0538}{0.05} \\\\v = 1.076 \ m/s[/tex]
The direction of the final velocity;
[tex]\theta = tan^{-1} (\frac{P_i_y}{P_i_x} )\\\\\theta = tan^{-1} (\frac{0.03 \times 1.2}{0.02 \times 2} )\\\\\theta = 42 \ ^0[/tex]
Thus, the speed of the resulting 50 g balls is 1.076 m/s and the direction is 42⁰.
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