0

A particle $P$, of mass $0.5$ kg, is moving with velocity $(4i+4j)$ m/s when it receives an impulse $I$ of magnitude $2.5$ Ns.

As a result of the impulse, the direction of motion of $P$ is deflected through an angle of $45^\circ$.

Given that $I=(\lambda i + \mu j)$ Ns, find all the possible pairs of $\lambda$ and $\mu$.

I've tried to draw a diagram to visualise the problem however each time i do this, I'm getting $\mu$ to be $0$.

From what I understand this is not possible, could someone point me in the right direction?

I've been stuck on this problem for an hour now.

tomi
  • 9,594
sdasd
  • 1
  • Hello @Priit and welcome. Please show us what you've done to get $\mu=0$. Then we can identify an error in your method or tell you that you are correct. Very few people will help you on this site without some evidence of effort on your part. – tomi May 04 '20 at 07:14
  • So, since it's deflected at 45 degrees. Then arctan(µ/λ ) = 0 because arctan(4/4) already = 45, Therefore µ = 0. @tomi – sdasd May 04 '20 at 07:16
  • @Priit the question is asking for the impulse vector, not the final velocity vector – Dhanvi Sreenivasan May 04 '20 at 07:19
  • Shouldn't this be in physics stack exchange – user600016 May 04 '20 at 07:53

2 Answers2

0

Based on your comment, what I can gather is that maybe the question was misinterpreted. Here, you are given that the final velocity is $45 ^o $ with respect to the initial velocity. This gives us two cases for the final velocity - either along x axis or y axis. I'll demonstrate one case here

Let final velocity be $v_f = a$i $$\textbf I = m\Delta\textbf v$$ $$\textbf I = 0.5(a \textbf i - (4\textbf i + 4\textbf j))$$ $$\textbf I = 0.5((a-4)\textbf i - 4\textbf j)$$

Now, using the fact that $|\textbf I| = 2.5 $ Ns, you can obtain $a$, resubstitute to get one pair of $(\lambda, \mu)$.

Repeat with $v_f = a \textbf j$

Dhanvi Sreenivasan
  • 5,637
0

Initial velocity is $u=4i+4j$. You have correctly identified this as travelling at an angle of $45^\circ$ to the horizontal.

Final velocity is either $v=ai$ (horizontal) or $v=bj$ (vertical). $a>0$ and $b>0$.

Consider the case where $v=ai$

Change in momentum is $m(v-u)$

$\lambda i + \mu j=0.5(ai-4i-4j)$

$\lambda i + \mu j=(0.5a-2)i-2j$

Vertical components: $\mu=-2$

Horizontal components: $\lambda=0.5a-2$

But $\lambda^2+\mu^2=2.5^2$

$(0.5a-2)^2=6.25-4$

$0.5a-2=\sqrt {2.25}$

$0.5a-2=1.5$ or $0.5a-2=-1.5$

$0.5a=3.5$ or $0.5a=0.5$

$a=7$ or $a=1$

Then $\lambda=3.5-2=1.5$ or $\lambda=0.5-2=-1.5$

Do similarly with the other case, where $v=bj$.

Change in momentum is $m(v-u)$

$\lambda i + \mu j=0.5(bj-4i-4j)$

$\lambda i + \mu j=-2i+(0.5b-2)j$

Vertical components: $\mu=0.5b-2$

Horizontal components: $\lambda=-2$

But $\lambda^2+\mu^2=2.5^2$

$(0.5b-2)^2=6.25-4$

$0.5b-2=\sqrt {2.25}$

$0.5b-2=1.5$ or $0.5b-2=-1.5$

$0.5b=3.5$ or $0.5b=0.5$

$b=7$ or $b=1$

Then $\mu=3.5-2=1.5$ or $\mu=0.5-2=-1.5$

tomi
  • 9,594