The potential energy (inJoules) of a particle of mass 1kgmoving in a plane is given byV= 3x+ 4y, the position coordinates of the point beingxandy, measured inmetres. If the particle is at rest at (6,4); thena)it crosses they-axis (x= 0) aty= –4b)its speed when it crosses they-axis is 10m/sc)its acceleration is of magnitude 5m/s2d)it moves in a straight line passing through the origin (0,0)Correct answer is option 'A,B,C'. Can you explain this answer?

Question

The potential energy (inJoules) of a particle of mass 1kgmoving in a plane is given byV= 3x+ 4y, the position coordinates of the point beingxandy, measured inmetres. If the particle is at rest at (6,4); thena)it crosses they-axis (x= 0) aty= &ndash;4b)its speed when it crosses they-axis is 10m/sc)its acceleration is of magnitude 5m/s2d)it moves in a straight line passing through the origin (0,0)Correct answer is option 'A,B,C'. Can you explain this answer?

Answer

To find the value of y when the particle crosses the y-axis (x=0), we need to set x=0 in the equation for potential energy:

V = 3x - 4y

0 = 3(0) - 4y

0 = -4y

This equation tells us that y=0 when x=0. However, since the particle is already at rest at the point (6,4), it has already crossed the y-axis at y=0. Therefore, there is no additional value of y when the particle crosses the y-axis.

Answer

Let at time 't' particle crosses y-axis

⇒ Particle crosses y-axis at y = –4
At (6, 4) U = 34 & KE = 0
At (0, 4); U = –16
⇒ KE = 50

⇒ v = 10 m/s while crossing y-axis
The correct answers are: its acceleration is of magnitude 5m/s², its speed when it crosses the y-axis is 10m/s, it crosses the y-axis (x = 0) at y = –4

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