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Chapter 31: Q. 2 (page 901)

A rocket cruises past a laboratory at 1.00x106m/sin the positive x-direction just as a proton is launched with velocity (in the laboratory frame) v=1.41x106i^+1.41x106j^m/s. What are the proton’s speed and its angle from the y-axis in (a) the laboratory frame and (b) the rocket frame?

Short Answer

Expert verified

(a) The speed of the proton is 1.99×106m/sat an angle of45∘ clockwise from the y-axis.
(b) The speed of the proton is 1.47×106m/sat an angle of 16.2∘clockwise from the y-axis.

Step by step solution

01

Part(a) Step1: Given information

We have given ,

Speed of rocket Vr=1.00x106m/s(along positive x-axis).

According to the question the speed of proton has two components:

Speed of proton localid="1649860951900" Vx=1.41x106m/s(along positive x-axis).

Speed of proton localid="1649860959788" Vy=1.41x106m/s(along positive y-axis).

We need to find the proton's speed and angle in laboratory frame.

02

Part(a) Step2: Simplify.

In the laboratory point of view all the velocities will be same as the question.

Magnitude of the proton velocity in frame of laboratory:

Vp=(Vx)2+(Vy)2

Vp=(1.41×106m/s)2+(1.41×106m/s)2Vp=1.99×106m/s

Clockwise angle from the y-axis:
tanθ=VxVytanθ=1.41×106m/s1.41×106m/stanθ=1θ=tan-1(1)θ=45∘



03

Part(b) Step1: Given information.

We have given,

Vx=1.41×106m/sVy=1.41×106m/s

We need to find the proton's speed and angle in rocket frame.

04

Part(b) Step2: Simplify.

In the rocket point of view the velocity of proton along y axis will be less as rocket also have some velocity along the positive x-axis:


Vx=(1.41×106-1.00×106)m/s=0.41×106m/sVy=1.41×106m/s

In this case the proton's speed will be:


Vp=(0.41×106)2+(1.41×106)2m/s


Vp=1.47×106m/s


Clockwise angle from the y-axis in frame of rocket:


tanθ=VxVy=(0.41×106)2m/s(1.41×106)2m/s=0.2908θ=tan-10.2908θ=16.2o


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