Answer :
Answer:
a) [tex]p_i=1.568\hat{i}+0.752 \hat{j}[/tex]
b) [tex]v_{fx}=1.668\ m.s^{-1}[/tex]
c) [tex]v_{fy}=0.7999\ m.s^{-1}[/tex]
Explanation:
Given masses:
[tex]m_1=0.49\ kg[/tex]
[tex]m_2=0.47\ kg[/tex]
Velocity of mass 1, [tex]v_1=3.2 \hat{i}\ m.s^{-1}[/tex]
Velocity of mass 2, [tex]v_2=1.6 \hat{j}\ m.s^{-1}[/tex]
a)
Initial momentum:
[tex]p_i=m_1.v_1+m_2.v_2[/tex]
[tex]p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}[/tex]
[tex]p_i=1.568\hat{i}+0.752 \hat{j}[/tex]
b)
magnitude of initial momentum:
[tex]p_i=\sqrt{1.568^2+0.752 ^2}[/tex]
[tex]p_i=1.739\ kg.m.s^{-1}[/tex]
From the conservation of momentum:
[tex]p_f=p_i[/tex]
[tex]m_f.v_f=1.739[/tex]
[tex]v_f=\frac{1.739}{0.49+0.47}[/tex]
[tex]v_f=1.85\ m.s^{-1}[/tex] is the magnitude of final velocity.
Direction of final velocity will be in the direction of momentum:
[tex]tan\theta=\frac{0.752 }{1.568}[/tex]
[tex]\theta=25.62^{\circ}[/tex]
[tex]\therefore v_{fx}=1.85\ cos25.62^{\circ}[/tex]
[tex]v_{fx}=1.668\ m.s^{-1}[/tex]
c)
Vertical component of final velocity:
[tex]v_{fy}=1.85\ sin 25.62^{\circ}[/tex]
[tex]v_{fy}=0.7999\ m.s^{-1}[/tex]
Answer:
The initial momentum of system is, [tex]1.56i+ 0.75j[/tex].
The horizontal component of final velocity is 0.780 m/s.
The vertical component of final velocity is 1.622 m/s.
Explanation:
Given data:
Mass of first ball is, [tex]m_{1}=0.49 \;\rm kg[/tex].
Mass of second ball is, [tex]m_{2}=0.47 \;\rm kg[/tex].
Initial velocity of ball 1 is, [tex]v_{1}=3.2 i \;\rm m/s[/tex].
Initial velocity of ball 2 is, [tex]v_{2}=1.6j \;\rm m/s[/tex].
(a)
The initial momentum of system is,
[tex]P_{i}=m_{1}v_{1}+m_{2}v_{2}\\P_{i}= (0.49 \times 3.2i)+ (0.47 \times 1.6j)\\P_{i}= 1.56i+ 0.75j[/tex]
Thus, initial momentum of system is, [tex]1.56i+ 0.75j[/tex].
(b)
Apply the conservation of linear momentum to obtain the value of final velocity as,
[tex]p_{i}=(m_{1}+m_{2})v\\\sqrt{1.56^{2}+0.75^{2}}=(0.49+0.47)v\\v=1.80 \;\rm m/s[/tex]
Then, the horizontal component of final velocity is,
[tex]v_{x}=vcos\theta[/tex]
Here, [tex]\theta[/tex] is the orientation of two masses. And its value is,
[tex]tan \theta = \dfrac{1.56}{0.75} \\\theta = tan^{-1}(\dfrac{1.56}{0.75}) \\\theta = 64.32 ^{\circ}[/tex]
Then,
[tex]v_{x}=vcos\theta\\v_{x}=1.80 \times cos64.32\\v_{x}=0.780 \;\rm m/s[/tex]
Thus, the horizontal component of final velocity is 0.780 m/s.
(c)
The vertical component of the final velocity
[tex]v_{y}=vsin\theta\\v_{y}=1.80 \times sin64.32\\v_{y}=1.622 \;\rm m/s[/tex]
Thus, the vertical component of final velocity is 1.622 m/s.
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