Solve This Physics Problem

Question

A ball of mass 197 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.5 s? (Do this problem by finding the components of the momentum first and then constructing the magnitude and direction of the momentum vector from the components. Find the magnitude in kg · m/s and the direction in degrees counterclockwise from the +x-axis. Assume the +x-axis points horizontally to the right.)

Answer 1

First, let's define how we'd be using the coordinate system: positive X-axis is along the horizontal so that the x-component of ball's starting velocity

v_xo=vcosθ=25m/ s * cos (30^o)=21.7m/s

Let's set vertical upward direction as the positive Y-axis . Therefore, the vertical component of starting velocity

v_yo=vsinθ=25m/ s * sin (30^o)=25m/ s * (1/2)=12.5m/s

would have a positive sign but the sign of (downward) acceleration of free fall would have negative

g=-9.8m/s²

The ball's horizontal component of velocity is not going to change in the course of ball's motion because the only force that would act on the ball is the vertical force of gravity F=mg

In the vertical direction, the ball would be move with a constant (negative) acceleration g starting with the velocity v_yo . At the time t=0.5 sec v_y=v_yo+gt= -9.8m/s²*0.5s=

12.5m/s-4.9m/s=7.6m/s

The absolute value of the velocity at t=0.5s therefore would be v=Sqrt(v_x²+v_y²)=Sqrt(7.6²+21.7²)m/s=22.3m/s

The ratio (v_y/v_x) would determine the velocity's angle α to the horizontal tan α=(v_y/v_x)=(7.6m/s)/(21.7m/s)==0.35 so

α=tan^-1(0.35)=13^o

The components and magnitude of momentum at 0.5s:

p_x=mv_x=0.197kg*21.7m/s=4.3kgm/7 ; p_y=0.197kg*7.6m/s=1.5kgm/s

The magnitude of vector of momentum

p=Sqrt(p_x²+p_y²)=4.6kgm/s

The angle of the vector p=mv to the horizon should be the same at that of the velocity vector v found above as 13^o

Answer 2

Hi Elle B.

To find the momentum of the ball after 0.5 seconds, we can first calculate the components of the momentum and then use vector addition to find the magnitude and direction of the momentum vector.

Given:

- Mass of the ball (m) = 197 g = 0.197 kg

- Initial velocity (u) = 25 m/s

- Launch angle (θ) = 30 degrees

- Time (t) = 0.5 seconds

We need to find the x-component (Px) and y-component (Py) of the momentum.

The x-component (Px) of the momentum is given by:

Px = m * vx

where vx is the horizontal component of velocity. We can find vx using the initial velocity and angle:

vx = u * cos(θ)

Px = 0.197 kg * 25 m/s * cos(30°)

Px = 0.197 kg * 25 m/s * √3/2

Px = 4.8875 kg·m/s

The y-component (Py) of the momentum is given by:

Py = m * vy

where vy is the vertical component of velocity. We can find vy using the initial velocity and angle:

vy = u * sin(θ)

Py = 0.197 kg * 25 m/s * sin(30°)

Py = 0.197 kg * 25 m/s * 0.5

Py = 2.4625 kg·m/s

Now, we have the x-component (Px) and y-component (Py) of the momentum. To find the magnitude and direction of the momentum vector, we can use the Pythagorean theorem and the inverse tangent function:

Magnitude (|P|) = √(Px^2 + Py^2)

|P| = √(4.8875^2 + 2.4625^2)

|P| ≈ 5.4607 kg·m/s (rounded to 4 decimal places)

Direction (θ) = atan(Py / Px)

θ = atan(2.4625 / 4.8875)

θ ≈ 27.0892 degrees counterclockwise from the +x-axis (rounded to 4 decimal places)

So, the magnitude of the momentum of the ball after 0.5 seconds is approximately 5.4607 kg·m/s, and the direction is approximately 27.0892 degrees counterclockwise from the +x-axis.

I hope this will help. I am happy to tutor you on any other questions you may have; please feel free to shoot me a message!