Atennis ball of mass m = 0.060 kg is traveling with a speed of 28 m/s and θ = 35 degrees. it strikes a wall and bounces off with a speed of 28 m/s. what is the magnitude of the force, in newtons, exerted on the ball if the collision time is 0.1 s? what direction was the force exerted by the wall?

19.15 Newton

Explanation:

Force on a body is rate of change of momentum.

Thus force on wall is =

 ΔP = P(final) - P(initial) [in vector]

Note : take component of velocity normal to wall (assuming wall smooth)

 P(final) = m*v*sin(Ф) [in negative x direction] = -0.06*28*sin(35)

 P(initial) = m*v*sin(Ф) [in positive x direction] = 0.06*28*sin(35)

Thus ΔP=2(0.06*28*sin(35))

Now ΔT is given =0.1 s

Force on wall = = =19.15 Newton

terd muffin is the correct answer

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s