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I. Multiple Choice Questions · Q1

Q.A particle of mass mm is moving with speed uu in a direction which makes 60∘60^\circ with respect to the x-axis. It undergoes an elastic collision with the wall. What is the change in momentum in the x and y direction?

(a) Δpx=−mu, Δpy=0\Delta p_x = -mu,\ \Delta p_y = 0
(b) Δpx=−2mu, Δpy=0\Delta p_x = -2mu,\ \Delta p_y = 0
(c) Δpx=0, Δpy=mu\Delta p_x = 0,\ \Delta p_y = mu
(d) Δpx=mu, Δpy=0\Delta p_x = mu,\ \Delta p_y = 0
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Step 1. Resolve the incoming velocity into components perpendicular and parallel to the wall. With the wall taken perpendicular to the x-axis, the x-component of the incoming velocity is ux=ucos⁡60∘=u/2u_x=u\cos60^\circ=u/2, and the y-component (along the wall) is uy=usin⁡60∘u_y=u\sin60^\circ.

Step 2. For an elastic collision with a fixed, rigid wall, only the component of velocity perpendicular to the wall reverses; the component parallel to the wall is unaffected (exactly the same rule used to derive the kinetic-theory pressure formula in Section 9.2.1). So after the collision, vx→−uxv_x\to-u_x and vy→uyv_y\to u_y (unchanged).

Step 3. Change in x-momentum: Δpx=m(−ux)−m(ux)=−2mux=−2m(u/2)=−mu\Delta p_x = m(-u_x) - m(u_x) = -2mu_x = -2m(u/2) = -mu.

Step 4. Change in y-momentum: Δpy=m(uy)−m(uy)=0\Delta p_y = m(u_y) - m(u_y) = 0, since the y-component never changes.

Step 5. Eliminating the others: (b) would be the change if the FULL speed uu (not its x-component) were reversed; (c) and (d) both incorrectly assign a nonzero change to Δpy\Delta p_y, which never changes in an elastic wall collision.

✓Final answer

(a) Δpx=−mu, Δpy=0\Delta p_x = -mu,\ \Delta p_y = 0.

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