Question

A particle of mass m is projected with velocity making an angle of 45∘ with the horizontal When the particle lands on the level ground the magnitude of the change in its momentum will be .

A. mv√2
B. zero
C. 2 mv
D. mv/√2

Answer 1

The magnitude of the change in momentum for a particle projected at 45° when it lands on level ground is mv√2, where m is the mass and v is the initial velocity.

Step-by-step explanation:

The question pertains to the conservation of momentum and involves calculating the change in momentum of a particle projected at an angle with the horizontal when it lands.

The particle's initial momentum vector has both horizontal and vertical components due to the 45° launch angle. When the particle lands, its vertical momentum component will have changed signs due to gravity, but the magnitude remains the same because it lands with the same speed it was launched (ignoring air resistance). However, the horizontal component remains unchanged.

The overall change in momentum, therefore, involves only the vertical components. Since momentum is a vector quantity, the change in the vertical component can be combined with the unchanged horizontal component to give a resultant vector that has a magnitude of mv√2, making option A the mentioned correct option in final answer.