IAT (IISER) 2025 | Work, Energy and Power Question 1 | Physics

A block of mass $M$ lies at rest connected to a massless spring of spring constant $k$ on a frictionless surface. A bullet of mass $m$ hits the block horizontally with speed $v$ as shown in the figure and is completely stuck to the block. What is the maximum compression in the spring resulting from this impact (assuming that at this point the spring is still not fully compressed)?

IAT (IISER) 2025 Physics - Work, Energy and Power Question 1 English

$\sqrt{\frac{m^2 v^2}{k(M+m)}}$

$$ \sqrt{\frac{m v^2}{k}} $$

$$ \sqrt{\frac{M v^2}{k}} $$

$$ \sqrt{\frac{m M v^2}{k(M+m)}} $$

IAT (IISER) 2025

MCQ (Single Correct Answer)

-1

A cart of mass $M$ is released from $A$, the highest point of a frictionless track, as shown in the figure. The cart travels along the track and enters the semicircular arc $D B C$ of radius $R$. The heights of the points $A$ and $B$ are $h_1$ and $h_2$ from the ground, respectively. Which of the following quantities does not play any role in ensuring that the cart does not leave the track?

IAT (IISER) 2025 Physics - Work, Energy and Power Question 2 English

$M$

$h_1$

$\quad h_2$

$R$

IAT (IISER) 2022

MCQ (Single Correct Answer)

-1

An object is released from rest from the inner edge of a hemispherical bowl, and it falls under gravity. The coefficient of kinetic friction between the object and the bowl is $\mu$. If the object covers an angular displacement $\theta$ with respect to the center of the hemisphere when it stops for the first time, which of the following expressions is correct?

$\mu=\cot (\theta)$

$\mu=\cot \left(\frac{\theta}{2}\right)$

$\mu=\tan (\theta)$

$\mu=\tan (\theta / 2)$

IAT (IISER) 2020

MCQ (Single Correct Answer)

-1

The potential energy of a point particle of mass $m$ undergoing rectilinear motion along the $x$-axis is given by

$$ V(x)=A x+B x^2 $$

What is the maximum speed attained by the particle if it starts from rest at $x=\frac{A}{B}$ ?

$\frac{3 A}{\sqrt{2 B m}}$

$\frac{A}{\sqrt{2 B m}}$

$\frac{2 A}{\sqrt{B m}}$

$A \sqrt{\frac{2}{B m}}$

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