Question

4-20 A hockey player receives a corner shot at a speed of
15 m/s at an angle of 30° with the v-axis and then shoots the ball
along the x-axis with a speed of 30 m/s. If the mass of the ball is
100 gm and it remains in contact with the hockey stick for 0.01s
the force imparted to the ball in the x-direction is:
600
Figure 4.105
(B) 187.5N
(D) 375 N
(A) 281.25 N
(C) 562.5 N​

Answer 1

Correct Question:

A hockey player receives a corner shot at a speed of 15 m/s at an angle of 30° with the y-axis and then shoots the ball along the x-axis with a speed of 30 m/s. If the mass of the ball is 100 gm and it remains in contact with the hockey stick for 0.01s the Force imparted to the ball in the x-direction is?

Answer:

• Force (F) Imparted is 375 N.

Given:

1. Angle made with y - axis (θ) = 30°
2. Mass of Ball (M) = 100
3. Time of contact (t) = 0.01 sec
4. Initial velocity of ball (u) = 15 m/s.
5. Final velocity of ball (v) = 30 m/s.

Explanation:

$\rule{300}{1.5}$

From Newton's second law of motion.

$\large{\boxed{\tt F = \dfrac{\Delta P}{t}}}$

$\bold{Here}\begin{cases}\text{F Denotes Force imparted} \\ \tt{\Delta} \text{P Denotes Change in momentum} \\ \text{t Denotes Time of contact}\end{cases}$

Now,

$\large{\boxed{\tt F = \dfrac{\Delta P}{t}}}$

Substituting the values,

$\longmapsto \large{\tt F = \dfrac{mv - mu}{t}}$

$\longmapsto \large{\tt F = \dfrac{m(v - u)}{t}}$

Now, Here as the Initial velocity is making an angle with the y - axis we need to resolve it into Components.

we get

• Along y - axis = u cos 30°
• Along x - axis = u sin 30°

As we need to Find the Force along x axis we need to consider only u sin 30°.

Substituting,

$\longmapsto \large{\tt F = \dfrac{m(v - [- u \sin 30 \degree])}{t}}$

Here we took u sin 30° has negative as its Direction is opposite to the final velocity (v)

$\longmapsto \large{\tt F = \dfrac{100 \: gm(30 - [- 15 \sin 30 \degree ])}{0.01}}$

∵ [1 gm = 10⁻³ Kg] Therefore,

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} \: Kg(30 - [- 15 \sin 30 \degree])}{0.01}}$

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} (30 - [- 15 \sin 30 \degree])}{0.01}}$

∵ sin 30° = ½

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} (30 - [- 15 \times \dfrac{1}{2}])}{0.01}}$

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} (30 - [- \dfrac{15}{2}])}{0.01}}$

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} (30 - [- 7.5])}{0.01}}$

$\longmapsto \large{\tt F = \dfrac{100 \times 10^{-3} (30 + 7.5)}{0.01}}$

$\longmapsto \large{\tt F = \dfrac{10^{-1} \times 37.5 }{0.01}}$

$\longmapsto \large{\tt F = \dfrac{10^{-1} \times 37.5 \times 100}{1}}$

$\longmapsto \large{\tt F = 37.5 \times 10^{2 - 1}}$

$\longmapsto \large{\tt F = 37.5 \times 10}$

$\longmapsto \large{\underline{\boxed{\red{\tt F = 375 \: N}}}}$

∴ Force (F) Imparted along x - axis is 375 N.

#Refer the attachment for figure.

Important Formulas:-

$\large \begin{tabular}{|c|c|c|} \cline{1 - 3} S.l No. & Quantities & Formula \\ \cline{1 - 3} 1 & Force & F = ma \\ \cline{1 - 3} 2 & Momentum & P = mv \\ \cline{1 - 3} 3 & Impulse & I = F.t \\ \cline{1 - 3} 4 & Weight & W = mg \\ \cline{1 - 3} 5 & Frictional Force & f = \mu .N \\ \cline{1 - 3} \end{tabular}$

$\rule{300}{1.5}$