Consider a rubber ball freely falling from a height h=4.9m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be :

Motion in A Straight Line PYQ — JEE Main 2009 · Shift 0 · Q60 with Solution

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Correct answer: B

Find the time to fall from height $h=4.9\text{ m}$

For free fall from rest, $h=\frac{1}{2}gt^2$ Given $h=4.9$ m and $g=9.8\text{ m/s}^2$ , $4.9=\frac{1}{2}(9.8)t^2=4.9t^2$ So, $t^2=1 \quad \Rightarrow \quad t=1\text{ s}$

Thus the ball hits the plate at $t=1\text{ s}$ .

Velocity just before collision

Taking upward as positive, the ball starts from rest and accelerates downward: $v=u-gt=0-9.8(1)=-9.8\text{ m/s}$

So just before hitting the plate, velocity is $-9.8\text{ m/s}$ .

Effect of totally elastic collision

Since the collision with the horizontal elastic plate is totally elastic and collision time is negligible, the speed remains same but direction reverses instantly.

Therefore, just after collision, $v=+9.8\text{ m/s}$

So the $v$ – $t$ graph is:

a straight line with slope $-g$ from $t=0$ to $t=1$ ,
then an instantaneous jump from $-9.8$ to $+9.8$ at $t=1$ ,
then again a straight line with slope $-g$ after that.

Equivalently,

\begin{cases} -gt, & 0\le t<1 \\ 2g-g t, & 1<t\le 2 \end{cases}$$ with a discontinuity at $t=1$. --- 4. **Height as a function of time** Let height above plate be $y$. Initially $y(0)=4.9$ m. For $0\le t\le 1$: $$y=4.9-\frac{1}{2}gt^2=4.9-4.9t^2$$ This decreases from $4.9$ to $0$. After the bounce, for $1\le t\le 2$, the ball rises with initial velocity $+9.8$ m/s from the plate: $$y=9.8(t-1)-\frac{1}{2}g(t-1)^2$$ $$y=9.8(t-1)-4.9(t-1)^2$$ This is a concave downward parabola, reaching back to $y=4.9$ m at $t=2$ s. So the $y$–$t$ graph is made of two downward-opening parabolic arcs meeting smoothly in position at $t=1$, but with slope changing sign abruptly there. --- 5. **Match with the correct option** Therefore the correct graph must show: - **velocity-time:** linear decrease, sudden jump upward at collision, then linear decrease again; - **height-time:** parabola down to zero at $t=1$, then parabola up after bounce. This corresponds to **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** My derived answer: **B** They agree.

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View written solutionFree

Correct answer: B

Find the time to fall from height $h=4.9\text{ m}$

For free fall from rest, $h=\frac{1}{2}gt^2$ Given $h=4.9$ m and $g=9.8\text{ m/s}^2$ , $4.9=\frac{1}{2}(9.8)t^2=4.9t^2$ So, $t^2=1 \quad \Rightarrow \quad t=1\text{ s}$

Thus the ball hits the plate at $t=1\text{ s}$ .

Velocity just before collision

Taking upward as positive, the ball starts from rest and accelerates downward: $v=u-gt=0-9.8(1)=-9.8\text{ m/s}$

So just before hitting the plate, velocity is $-9.8\text{ m/s}$ .

Effect of totally elastic collision

Since the collision with the horizontal elastic plate is totally elastic and collision time is negligible, the speed remains same but direction reverses instantly.

Therefore, just after collision, $v=+9.8\text{ m/s}$

So the $v$ – $t$ graph is:

a straight line with slope $-g$ from $t=0$ to $t=1$ ,
then an instantaneous jump from $-9.8$ to $+9.8$ at $t=1$ ,
then again a straight line with slope $-g$ after that.

Equivalently,

\begin{cases} -gt, & 0\le t<1 \\ 2g-g t, & 1<t\le 2 \end{cases}$$ with a discontinuity at $t=1$. --- 4. **Height as a function of time** Let height above plate be $y$. Initially $y(0)=4.9$ m. For $0\le t\le 1$: $$y=4.9-\frac{1}{2}gt^2=4.9-4.9t^2$$ This decreases from $4.9$ to $0$. After the bounce, for $1\le t\le 2$, the ball rises with initial velocity $+9.8$ m/s from the plate: $$y=9.8(t-1)-\frac{1}{2}g(t-1)^2$$ $$y=9.8(t-1)-4.9(t-1)^2$$ This is a concave downward parabola, reaching back to $y=4.9$ m at $t=2$ s. So the $y$–$t$ graph is made of two downward-opening parabolic arcs meeting smoothly in position at $t=1$, but with slope changing sign abruptly there. --- 5. **Match with the correct option** Therefore the correct graph must show: - **velocity-time:** linear decrease, sudden jump upward at collision, then linear decrease again; - **height-time:** parabola down to zero at $t=1$, then parabola up after bounce. This corresponds to **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** My derived answer: **B** They agree.

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