Q.A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time. (Take acceleration in the backward direction as positive).

Concept: Impulsive Acceleration (Newton's Second Law during a collision) — the ball's acceleration is zero except during the very short bat-ball contact, when a large force produces a large, brief acceleration spike.

The ball moves uniformly forward before the hit, so its acceleration is zero. During the very short bat-ball contact, a large force acts in the backward direction, producing a large positive acceleration spike (since backward is taken as positive). After the hit, the ball moves uniformly backward, so acceleration returns to zero.

1. Before $t = 0$: ball moves forward with constant velocity → $a = 0$. …

Why this approach works

The problem asks for the acceleration vs time graph of a cricket ball that is hit by a bat. The key is to recognise that the ball is under no net force except during the extremely brief collision with the bat. Before the hit, it moves uniformly (constant velocity, so zero acceleration). After the hit, it again moves uniformly in the opposite direction (again zero acceleration). The only acceleration occurs during the impact itself.

Since the bat exerts a large force for a very short time, the acceleration is correspondingly large and brief. We are told to take the backward direction (the direction of the force from the bat) as positive. So the acceleration during the hit is a large positive spike.

Step-by-step reasoning

1. Before the hit

   The ball moves with constant velocity toward the bat. Since velocity is constant, acceleration is zero. On the $a$-$t$ graph, this is a flat line at $a = 0$ for all $t$ before the collision.

2. During the hit

   The bat exerts a force on the ball for a very short time $\Delta t$. This force is large and roughly constant in magnitude (idealised as a rectangular pulse). The acceleration is $a = F/m$, where $F$ is the force from the bat. Since the bat pushes the ball backward (the positive direction), $a$ is positive and large. On the graph, this appears as a tall rectangle of height $a_{\text{max}}$ and width $\Delta t$.

3. After the hit

   The ball leaves the bat with a new constant velocity in the backward direction. Again, constant velocity means zero acceleration. So the graph returns to $a = 0$ for all $t$ after the collision.

4. The shape of the pulse

   In reality, the force varies smoothly, but for a "very short time interval" we idealise it as constant during contact. The area under the $a$-$t$ curve equals the change in velocity:

where $\Delta v$ is the change in velocity (from forward to backward). This area is fixed by the ball's speed change; a shorter $\Delta t$ means a taller $a_{\text{max}}$. …

Concept: Deriving the Spike's Shape from the Impulse-Momentum Theorem, Not from "the Force is Large"

Method: Work Backward from the Required $\Delta v$ (impulse-momentum theorem fixes the graph's area, which then forces its height)

The stored answer reasons forward — "the bat exerts a large force, so the acceleration is large." This method reverses the logic: it first pins down the exact velocity change the ball must undergo (a fact known from before/after the hit, with no reference to the force at all), and then uses the impulse-momentum theorem to show that a fixed, non-zero area under the $a$-$t$ curve, squeezed into a vanishingly short base, is mathematically forced to have a very large height — deriving the spike's shape as a consequence, rather than assuming it from "a bat hits hard."

Setup

Let the ball's forward speed (before the hit) have magnitude $u$. Backward is taken as positive, so before the hit $v_{\text{before}} = -u$ (forward = negative in this convention); after an idealised elastic-type reversal, $v_{\text{after}} = +u$ (now moving backward, positive).

Steps

1. Before the hit — zero acceleration, from Newton's first law directly. The ball moves with constant velocity $-u$ (no force acts on it in flight, ignoring air resistance) — by Newton's first law, constant velocity is only possible if the net force, and hence the acceleration, is exactly zero: $a=0$ for all $t<0$.

2. After the hit — same reasoning, zero acceleration again. Once contact ends, the ball again moves at a new constant velocity $+u$ — by the same first-law argument, $a=0$ for all $t>\Delta t$ (where $\Delta t$ is the brief contact duration).

3. Compute the REQUIRED velocity change — this does not depend on the force at all. From Steps 1–2:

This number is fixed by the before/after speeds alone.

4. Invoke the impulse–momentum theorem to relate this $\Delta v$ to the acceleration graph's area:

The area under the $a$-$t$ curve, over the contact interval, must equal exactly $2u$ — a fixed, non-zero number — regardless of how short $\Delta t$ is.

5. Deduce the spike's height from its area and (very short) base. Area $=$ base $\times$ average height, so

Since $\Delta t$ is "very short" by the problem's own wording, this ratio is necessarily large — the shorter the contact, the taller the spike must be, purely to keep the area (which is fixed at $2u$ by Step 3) unchanged. This is the mathematical reason the graph must show a tall, narrow pulse — not merely a description of "a big force acting briefly." …