Question

Please answer Q.4 in image

Answer 1

Assume right as i direction and upward as j direction
mass = 100g = 0.1kg

initial velocity =5(sin60 i + cos60 j) = 5(√3/2i + 1/2j) m/s
initial momentum = 0.1×5(√3/2i + 1/2j) kgm/s = 0.5(√3/2i + 1/2j) kgm/s

final velocity = 5(sin60 (-i) + cos60 j) = 5(-√3/2i + 1/2j) m/s
final momentum = 0.1×5(-√3/2i + 1/2j) kgm/s = 0.5(-√3/2i + 1/2j) kgm/s

change in momentum = 0.5(-√3/2i + 1/2j) - 0.5(√3/2i + 1/2j)
                                = -√3/4i + 1/4j -√3/4i - 1/4j
                                = -√3/4i -√3/4i =  -√3/2i kgm/s
time taken = 2×10⁻³ s

force = change in momentum/time
        = -√3/2 / 2×10⁻³ i N
        = -1000√3/4 i N
        = -250√3 i N
since force is negative, direction is opposite of i or direction is to the left.
Answer is C.

Answer 2

Answer-

(4) (C) 250√3 N to left

Solution:

To find the force,

Initial momentum,$P_{1}$=mvsinΘi + mvcosΘj

Final momentum,$P_{2}$ =−mvsinΘi +bmvcosΘj

Where,

$P_{1} ,P_{2}$ − Initial and final momentum

m - mass

v - speed

Therefore,

Force = Change in momentum / Change in time

ΔP/ΔT

−2mvsinΘ/2× ${10}^{ - 3}$

Substituting the values,

We get,

−2(0.1)(5) sin60/2× ${10}^{ - 3}$

Force =−250√3N

The negative sign indicated the opposite direction of the force.

Hence, the force applied by the wall is 250√3N to left.

Hence,

option C is correct